noi2016解题报告

D1T1：
首先转化成统计AA型字符串有几种。
st[i]表示从i位置开始的AA型字符串有几个，ed[i]表示到i结束的有几个。
ans=∑st[i]*ed[i-1]
然后枚举A的长度L（AA长度的一半），i=k*L,j=(k+1)*L
观察x=lcp(i,j)和y=lcs(i-1,j-1)发现只有当x+y>=L时存在长度为L的AA型，然后显然是连续的一段，算算从哪开始到哪结束，差分一下，最后求和就好了。
lcp可以用hash或者kmp（hash有时候要卡常数）

#include <bits/stdc++.h>#define ll long long#define N 30009#define A 131#define mod 998244353#define mid (l+r>>1)#define pd(x) ((x)<0?(x+mod):x)using namespace std;ll n,st[N],ed[N],h[N],H[N],p[N];char str[N];inline ll get(ll x,ll len){    return pd(h[x+len-1]-h[x-1]*p[len]%mod);}inline ll lcp(ll i,ll j){    ll l=1,r=(n-j+1),ans=0;    while (l<=r)    {        if (get(i,mid)==get(j,mid)) ans=mid,l=mid+1;        else r=mid-1;    }    return ans;}inline ll Get(ll x,ll len){    return pd(H[x-len+1]-H[x+1]*p[len]%mod);}inline ll lcs(ll i,ll j){    ll l=1,r=i,ans=0;    while (l<=r)    {        if (Get(i,mid)==Get(j,mid)) ans=mid,l=mid+1;        else r=mid-1;    }    return ans;}int main(){    ll T;    scanf("%lld",&T);    while (T--)    {        scanf("%s",str+1);        n=strlen(str+1);        for (ll i=0;i<=n+1;i++) st[i]=ed[i]=0;        p[0]=1,h[0]=H[0]=h[n+1]=H[n+1]=0;        for (ll i=1;i<=n;i++) p[i]=p[i-1]*A%mod;        for (ll i=1;i<=n;i++) h[i]=(h[i-1]*A+str[i]-'a'+1)%mod;        for (ll i=n;i>=1;i--) H[i]=(H[i+1]*A+str[i]-'a'+1)%mod;        for (ll L=1;L<=(n>>1);L++)            for (ll k=1;(k+1)*L<=n;k++)            {                ll i=k*L,j=i+L;                ll x=min(lcp(i,j),L),y=min(lcs(i-1,j-1),L-1);                if (x+y>=L)                {                    ll t=x+y-L+1;                    st[i-y]++;                    st[i-y+t]--;                    ed[j+x-1]++;                    ed[j+x-1-t]--;                }            }        for (ll i=1;i<=n;i++) st[i]+=st[i-1];        for (ll i=n;i>=1;i--) ed[i]+=ed[i+1];        ll ans=0;        for (ll i=1;i<n;i++)            ans+=st[i+1]*ed[i];        printf("%lld\n",ans);    }    return 0;}

D1T2：
仔细撕烤后发现答案<=2，-1情况肯定要么c>=n*m-1,要么c==n*m-2且剩下两只连在一起。
然后答案为零就是原来就有两块及以上，也就是不完全联通。
答案等于1就是完全联通但是存在割点。否则答案等于2。
这样我们想到了tarjan求割点跟强连通分量。
首先将蛐蛐的联通情况求出来。
然后每个点的5*5的方阵内的点都是有希望的。然后对于这些点中不属于蛐蛐的点跑一遍tarjan，算出来的割点如果有满足旁边3*3的方阵中确实有蛐蛐的才是真的割点。（画个图自己想想就好啦）。整幅图不连通的话，必然有一个蛐蛐的周围24个格子属于两个或以上的联通块。然后根据这个判断联通情况就好了。

#include <bits/stdc++.h>#define gc getchar()#define N 100009#define mp make_pair#define pb push_back#define pa pair<int,int>#define ll long longusing namespace std;const int sed=195337;struct hash{    #define mod 100007    #define M 5000010    int head[mod],dh[mod];    int x[M],y[M],w[M],next[M],siz,cnt;    inline void clear()    {        ++cnt,siz=0;    }    void ins(int _x,int _y,int _i)    {        int s=(1LL*_x*sed+_y)%mod;        if (dh[s]!=cnt) dh[s]=cnt,head[s]=0;        next[++siz]=head[s],head[s]=siz;        x[siz]=_x,y[siz]=_y,w[siz]=_i;    }    int count(int _x,int _y)    {        int s=(1LL*_x*sed+_y)%mod;        if (dh[s]!=cnt) return -1;        for (s=head[s];s;s=next[s])            if (x[s]==_x&&y[s]==_y) return w[s];        return -1;    }}pos,Pos;const int TT=24;const int dx[8]={0,0,1,-1,1,-1,1,-1};const int dy[8]={1,-1,0,0,-1,1,1,-1};int n,m,c,cnt,flag,Bel,bel[N],Be,be[N*TT],dfn[N*TT],low[N*TT],num;bool is_cut[N*TT];pa p[N],q[N*TT];vector<int> go[N],Go[N*TT],point[N];int read(){    char ch;    int x=1;    while (ch=gc,ch<'0'||ch>'9') if (ch=='-') x=-1;    int s=ch-'0';    while (ch=gc,ch>='0'&&ch<='9') s=s*10+ch-'0';    return s*x;}bool check(int x){    for (int i=0;i<8;i++)        if (pos.count(q[x].first+dx[i],q[x].second+dy[i])!=-1)            return 1;    return 0;}void dfs(int x){    bel[x]=Bel;    for (int i=0;i<(int)go[x].size();i++)        if (!bel[go[x][i]]) dfs(go[x][i]);}void tarjan(int x,int fa){    int son=0;    low[x]=dfn[x]=++cnt,be[x]=Be;    for (int i=0;i<Go[x].size();i++)    {        int to=Go[x][i];        if (to!=fa)        {            if (!dfn[to])            {                ++son,tarjan(to,x);                low[x]=min(low[x],low[to]);                if (low[to]>=dfn[x]) is_cut[x]=1;            }            else low[x]=min(low[x],dfn[to]);        }    }    if (fa==-1&&son==1) is_cut[x]=0;}inline bool Not_Connected(){    for (int i=1;i<=c;i++)        for (int j=0;j<8;j++)        {            int now=pos.count(p[i].first+dx[j],p[i].second+dy[j]);            if (now!=-1) go[i].pb(now);        }    num=cnt=Bel=Be=0;    for (int i=1;i<=c;i++)        if (!bel[i]) ++Bel,dfs(i);    for (int i=1;i<=c;i++)        for (int x=-2;x<=2;x++)            for (int y=-2;y<=2;y++)                if (x!=0||y!=0)                {                    pa g=mp(p[i].first+x,p[i].second+y);                    if (g.first<1||g.first>n||g.second<1||g.second>m)                        continue;                    if (pos.count(p[i].first+x,p[i].second+y)==-1&&Pos.count(p[i].first+x,p[i].second+y)==-1)                    {                        q[++num]=g,Pos.ins(p[i].first+x,p[i].second+y,num);                        point[bel[i]].pb(num);                    }                }    if (num==TT*c) return 0;    for (int i=1;i<=num;i++)        for (int j=0;j<4;j++)        {            int now=Pos.count(q[i].first+dx[j],q[i].second+dy[j]);            if (now!=-1) Go[i].pb(now);        }    for (int i=1;i<=num;i++)        if (!dfn[i]) ++Be,tarjan(i,-1);    for (int i=1;i<=num;i++)        if (is_cut[i]&&check(i)) flag=1;    for (int i=1;i<=Bel;i++)    {        int Num=-1;        for (int j=0;j<point[i].size();j++)            if (Num==-1) Num=be[point[i][j]];            else if (be[point[i][j]]!=Num) return 1;    }    return 0;}inline void CL(){    for (int i=1;i<=num;i++)        Go[i].clear(),is_cut[i]=dfn[i]=0;    for (int i=1;i<=c;i++)        point[i].clear(),go[i].clear(),bel[i]=0;    pos.clear(),Pos.clear();}int main(){    int T=read();    while (T--)    {        n=read(),m=read(),c=read();        for (int i=1;i<=c;i++)            p[i].first=read(),p[i].second=read();        if (c>=(ll)n*m-1)        {            puts("-1");            continue;        }        int ans=2;        flag=(n==1)|(m==1);        for (int i=1;i<=c;i++) pos.ins(p[i].first,p[i].second,i);        if (Not_Connected())        {            puts("0");            CL();            continue;        }        if ((ll)n*m==c+2)        {            if (Be==1||!c) puts("-1");            else puts("0");            CL();            continue;        }        puts(flag?"1":"2");        CL();    }    return 0;}

D1T3：
有趣的数论题。。感觉自己数论还是太差啦。。。
考虑一个数是否为纯循环小数。
如果x是的话，必然有x−⌊x⌋=x⋅kp−⌊x⋅kp⌋(p∈N+)
因为⌊x⋅kp⌋−⌊x⌋∈Z
所以x⋅kp−x∈Z
设x=ab(a,b∈N+)
所以b|kp−1
即kp≡1modb
由欧拉定理，当k,b互质时，有kφ(b)≡1modb
所以只要k,b互质即可
那么原题也就是

Ans=∑i=1n∑j=1m[(i,j)=1][(j,k)=1]=∑j=1m[(j,k)=1]∑d|jμ(d)⌊nd⌋=∑d=1min(n,m)μ(d)⌊nd⌋∑j=1⌊md⌋[(j⋅d,k)=1]=∑d=1min(n,m)[(d,k)=1]μ(d)⌊nd⌋∑j=1⌊md⌋[(j,k)=1]

设f(n)=∑ni=1[(i,k)=1]
发现f(n)=⌊nk⌋f(k)+f(nmodk)
（应该挺显然的）
然后可以预处理出n<=k的f(n,k)，就可以o(1)求它的前缀和，就可以轻松分块啦。

Ans=∑d=1min(n,m)[(d,k)=1]μ(d)⌊nd⌋f(⌊md⌋)

然后就需要快速算出g(n,k)=∑ni=1[(i,k)=1]μ(i)
的前缀和啦。
设n=pc⋅q(p∈P)
那么g(n,k)=∑ni=1[(i,q)=1]μ(i)−∑⌊np⌋i=1[(i⋅p,q)=1]μ(i⋅p)
因为所有跟q互质且跟n不互质的数a均有a=px⋅y(x>=1)
而x>=1时μ(a)=0
所以只要考虑x=1的情况。
则g(n,k)=g(n,q)−∑⌊np⌋i=1[(i,q)=1][(i,p)=1]μ(i)⋅μ(p)
因为(i,p)不等于1时μ(p⋅i)=0
所以g(n,k)=g(n,q)−μ(p)∑⌊np⌋i=1[(i,k)=1]μ(i)=g(n,q)−μ(p)g(⌊np⌋,k)=g(n,q)+g(⌊np⌋,k)
然后我们想到了递归求解，显然递归层数小于等于n的质因子个数。
然后考虑边界情况：当n=0时，g(n,k)=0;
当k=1时，g(n,k)=g(n,1)=∑ni=1μ(i)
这显然可以杜教筛求解。
注意对杜教筛和递归都用上记忆化搜索（map）就可以轻易艹过去啦。

#include <bits/stdc++.h>#define gc getchar()#define ll long long#define N 2009#define M 1000009 using namespace std;ll n,m,k,f[N],P[M],n_now,pri[M],cnt,mu[M],sum[M];bool pd[M];vector<ll> q;map<ll,ll> mp;map<pair<ll,ll>,ll> Mp;ll get_mu(ll x){    return (x<M)?sum[x]:mp[x];}ll get(ll n){    if (n<M||mp.count(n)) return get_mu(n);    ll ans=1;    for (ll i=2,j;i<=n;i=j+1)    {        j=n/(n/i),get(n/i);        ans-=(j-i+1)*get_mu(n/i);    }    return mp[n]=ans;}ll g(ll n,ll k){    //g(n,k)=g(n,q)+g(n/p,k);    if (!k) return get(n);    if (n<=1) return n;    if (Mp.count(make_pair(n,k))) return Mp[make_pair(n,k)];    return Mp[make_pair(n,k)]=g(n,k-1)+g(n/q[k-1],k);}int main(){    memset(pd,0,sizeof(pd));    mu[1]=pd[1]=1;    for (ll i=2;i<M;i++)    {        if (!pd[i]) pri[++cnt]=i,mu[i]=-1;        for (ll j=1;j<=cnt&&pri[j]*i<M;j++)        {            pd[pri[j]*i]=1;            if (i%pri[j]==0)            {                mu[i*pri[j]]=0;                break;            }            mu[i*pri[j]]=-mu[i];        }    }    for (ll i=1;i<M;i++) sum[i]=sum[i-1]+mu[i];    scanf("%lld%lld%lld",&n,&m,&k);    for (int i=1;pri[i]<=k;i++)        if (k%pri[i]==0) q.push_back(pri[i]);    n_now=n;    for (ll i=1;i<=k;i++) f[i]=f[i-1]+(__gcd(i,k)==1ll);    ll ans=0;    for (ll i=1,j;i<=min(n,m);i=j+1)    {        j=min(n/(n/i),m/(m/i));        ans+=((m/i/k)*f[k]+f[m/i%k])*(n/i)*(g(j,q.size())-g(i-1,q.size()));    }    printf("%lld\n",ans);    return 0;}

D2T1：
按长度从大到小排序，对x，y离散化。
从大到小插入线段，原问题可以变为一段区间上的(r-l+1)条线段覆盖的点中有一个点被覆盖大于等于m次。
然后每次插入一条线段，不断拿掉现存的最长的线段（保证最大覆盖次数大于等于m），然后加入一条更短的线段后，显然如果对 上一条长线段从l开始是成立的，对它肯定也是成立的，即l肯定是单调递增的，所以类似单调队列的方式搞一搞就好了。

#include <iostream>#include <cstdio>#include <algorithm>#define inf 0x7fffffff#define gc (*buf++)#define N 500009using namespace std;int n,m,b[N<<1],c[N<<1],d[N<<1],Max[N<<3],add[N<<3],l=1,r=0,ans=inf,num;char Buf[30000000],*buf=Buf;struct seg{    int x,y,len;    inline bool operator <(const seg& rhs) const    {        return len>rhs.len;    }}a[N];inline int read(){    char ch;    int x=1;    while (ch=gc,ch<'0'||ch>'9') if (ch=='-') x=-1;    int s=ch-'0';    while (ch=gc,ch>='0'&&ch<='9') s=s*10+ch-48;    return s*x;}inline void ins(int l,int r,int L,int R,int k,int x){    if (L<=l&&R>=r)    {        add[k]+=x,Max[k]+=x;        return;    }    int mid=l+r>>1;    if (L<=mid) ins(l,mid,L,R,k<<1,x);    if (R>mid) ins(mid+1,r,L,R,k<<1|1,x);    Max[k]=max(Max[k<<1],Max[k<<1|1]);    Max[k]+=add[k];}int low(int x){    int l=1,r=num,mid;    while (l<r)    {        mid=(l+r)>>1;        if (b[mid]>=x) r=mid;        else l=mid+1;    }    return l;}int main(){    fread(Buf,1,30000000,stdin);    n=read(),m=read();    num=0;    for (int i=1;i<=n;i++)    {        a[i].x=read();a[i].y=read();        a[i].len=a[i].y-a[i].x;        b[++num]=a[i].x;        b[++num]=a[i].y;    }    sort(a+1,a+n+1);    sort(b+1,b+num+1);    for (int i=1;i<=n;i++) a[i].x=low(a[i].x),a[i].y=low(a[i].y);    for (int r=1;r<=n;r++)    {        ins(1,num,a[r].x,a[r].y,1,1);        while (Max[1]>=m)        {            ans=min(ans,a[l].len-a[r].len);            ins(1,num,a[l].x,a[l].y,1,-1);            l++;        }    }    if (ans-inf) printf("%d\n",ans); else puts("-1");    return 0;}

D2T2：
结论太TM多了。要不扔课件跑（丢链接跑）。。？

https://wenku.baidu.com/view/7842de6784868762cbaed52e.html

// This is an empty program with decimal lib#include <cstdlib>#include <cstring>#include <string>// ---------- decimal lib start ----------const int PREC = 3100;class Decimal {    public:        Decimal();        Decimal(const std::string &s);        Decimal(const char *s);        Decimal(int x);        Decimal(long long x);        Decimal(double x);        bool is_zero() const;        // p (p > 0) is the number of digits after the decimal point        std::string to_string(int p) const;        double to_double() const;        friend Decimal operator + (const Decimal &a, const Decimal &b);        friend Decimal operator + (const Decimal &a, int x);        friend Decimal operator + (int x, const Decimal &a);        friend Decimal operator + (const Decimal &a, long long x);        friend Decimal operator + (long long x, const Decimal &a);        friend Decimal operator + (const Decimal &a, double x);        friend Decimal operator + (double x, const Decimal &a);        friend Decimal operator - (const Decimal &a, const Decimal &b);        friend Decimal operator - (const Decimal &a, int x);        friend Decimal operator - (int x, const Decimal &a);        friend Decimal operator - (const Decimal &a, long long x);        friend Decimal operator - (long long x, const Decimal &a);        friend Decimal operator - (const Decimal &a, double x);        friend Decimal operator - (double x, const Decimal &a);        friend Decimal operator * (const Decimal &a, int x);        friend Decimal operator * (int x, const Decimal &a);        friend Decimal operator / (const Decimal &a, int x);        friend bool operator < (const Decimal &a, const Decimal &b);        friend bool operator > (const Decimal &a, const Decimal &b);        friend bool operator <= (const Decimal &a, const Decimal &b);        friend bool operator >= (const Decimal &a, const Decimal &b);        friend bool operator == (const Decimal &a, const Decimal &b);        friend bool operator != (const Decimal &a, const Decimal &b);        Decimal & operator += (int x);        Decimal & operator += (long long x);        Decimal & operator += (double x);        Decimal & operator += (const Decimal &b);        Decimal & operator -= (int x);        Decimal & operator -= (long long x);        Decimal & operator -= (double x);        Decimal & operator -= (const Decimal &b);        Decimal & operator *= (int x);        Decimal & operator /= (int x);        friend Decimal operator - (const Decimal &a);        // These can't be called        friend Decimal operator * (const Decimal &a, double x);        friend Decimal operator * (double x, const Decimal &a);        friend Decimal operator / (const Decimal &a, double x);        Decimal & operator *= (double x);        Decimal & operator /= (double x);    private:        static const int len = PREC / 9 + 1;        static const int mo = 1000000000;        static void append_to_string(std::string &s, long long x);        bool is_neg;        long long integer;        int data[len];        void init_zero();        void init(const char *s);};Decimal::Decimal() {    this->init_zero();}Decimal::Decimal(const char *s) {    this->init(s);}Decimal::Decimal(const std::string &s) {    this->init(s.c_str());}Decimal::Decimal(int x) {    this->init_zero();    if (x < 0) {        is_neg = true;        x = -x;    }    integer = x;}Decimal::Decimal(long long x) {    this->init_zero();    if (x < 0) {        is_neg = true;        x = -x;    }    integer = x;}Decimal::Decimal(double x) {    this->init_zero();    if (x < 0) {        is_neg = true;        x = -x;    }    integer = (long long)x;    x -= integer;    for (int i = 0; i < len; i++) {        x *= mo;        if (x < 0) x = 0;        data[i] = (int)x;        x -= data[i];    }}void Decimal::init_zero() {    is_neg = false;    integer = 0;    memset(data, 0, len * sizeof(int));}bool Decimal::is_zero() const {    if (integer) return false;    for (int i = 0; i < len; i++) {        if (data[i]) return false;    }    return true;}void Decimal::init(const char *s) {    this->init_zero();    is_neg = false;    integer = 0;    // find the first digit or the negative sign    while (*s != 0) {        if (*s == '-') {            is_neg = true;            ++s;            break;        } else if (*s >= 48 && *s <= 57) {            break;        }        ++s;    }    // read the integer part    while (*s >= 48 && *s <= 57) {        integer = integer * 10 + *s - 48;        ++s;    }    // read the decimal part    if (*s == '.') {        int pos = 0;        int x = mo / 10;        ++s;        while (pos < len && *s >= 48 && *s <= 57) {            data[pos] += (*s - 48) * x;            ++s;            x /= 10;            if (x == 0) {                ++pos;                x = mo / 10;            }        }    }}void Decimal::append_to_string(std::string &s, long long x) {    if (x == 0) {        s.append(1, 48);        return;    }    char _[30];    int cnt = 0;    while (x) {        _[cnt++] = x % 10;        x /= 10;    }    while (cnt--) {        s.append(1, _[cnt] + 48);    }}std::string Decimal::to_string(int p) const {    std::string ret;    if (is_neg && !this->is_zero()) {        ret = "-";    }    append_to_string(ret, this->integer);    ret.append(1, '.');    for (int i = 0; i < len; i++) {        // append data[i] as "%09d"        int x = mo / 10;        int tmp = data[i];        while (x) {            ret.append(1, 48 + tmp / x);            tmp %= x;            x /= 10;            if (--p == 0) {                break;            }        }        if (p == 0) break;    }    if (p > 0) {        ret.append(p, '0');    }    return ret;}double Decimal::to_double() const {    double ret = integer;    double k = 1.0;    for (int i = 0; i < len; i++) {        k /= mo;        ret += k * data[i];    }    if (is_neg) {        ret = -ret;    }    return ret;}bool operator < (const Decimal &a, const Decimal &b) {    if (a.is_neg != b.is_neg) {        return a.is_neg && (!a.is_zero() || !b.is_zero());    } else if (!a.is_neg) {        // a, b >= 0        if (a.integer != b.integer) {            return a.integer < b.integer;        }        for (int i = 0; i < Decimal::len; i++) {            if (a.data[i] != b.data[i]) {                return a.data[i] < b.data[i];            }        }        return false;    } else {        // a, b <= 0        if (a.integer != b.integer) {            return a.integer > b.integer;        }        for (int i = 0; i < Decimal::len; i++) {            if (a.data[i] != b.data[i]) {                return a.data[i] > b.data[i];            }        }        return false;    }}bool operator > (const Decimal &a, const Decimal &b) {    if (a.is_neg != b.is_neg) {        return !a.is_neg && (!a.is_zero() || !b.is_zero());    } else if (!a.is_neg) {        // a, b >= 0        if (a.integer != b.integer) {            return a.integer > b.integer;        }        for (int i = 0; i < Decimal::len; i++) {            if (a.data[i] != b.data[i]) {                return a.data[i] > b.data[i];            }        }        return false;    } else {        // a, b <= 0        if (a.integer != b.integer) {            return a.integer < b.integer;        }        for (int i = 0; i < Decimal::len; i++) {            if (a.data[i] != b.data[i]) {                return a.data[i] < b.data[i];            }        }        return false;    }}bool operator <= (const Decimal &a, const Decimal &b) {    if (a.is_neg != b.is_neg) {        return a.is_neg || (a.is_zero() && b.is_zero());    } else if (!a.is_neg) {        // a, b >= 0        if (a.integer != b.integer) {            return a.integer < b.integer;        }        for (int i = 0; i < Decimal::len; i++) {            if (a.data[i] != b.data[i]) {                return a.data[i] < b.data[i];            }        }        return true;    } else {        // a, b <= 0        if (a.integer != b.integer) {            return a.integer > b.integer;        }        for (int i = 0; i < Decimal::len; i++) {            if (a.data[i] != b.data[i]) {                return a.data[i] > b.data[i];            }        }        return true;    }}bool operator >= (const Decimal &a, const Decimal &b) {    if (a.is_neg != b.is_neg) {        return !a.is_neg || (a.is_zero() && b.is_zero());    } else if (!a.is_neg) {        // a, b >= 0        if (a.integer != b.integer) {            return a.integer > b.integer;        }        for (int i = 0; i < Decimal::len; i++) {            if (a.data[i] != b.data[i]) {                return a.data[i] > b.data[i];            }        }        return true;    } else {        // a, b <= 0        if (a.integer != b.integer) {            return a.integer < b.integer;        }        for (int i = 0; i < Decimal::len; i++) {            if (a.data[i] != b.data[i]) {                return a.data[i] < b.data[i];            }        }        return true;    }}bool operator == (const Decimal &a, const Decimal &b) {    if (a.is_zero() && b.is_zero()) return true;    if (a.is_neg != b.is_neg) return false;    if (a.integer != b.integer) return false;    for (int i = 0; i < Decimal::len; i++) {        if (a.data[i] != b.data[i]) return false;    }    return true;}bool operator != (const Decimal &a, const Decimal &b) {    return !(a == b);}Decimal & Decimal::operator += (long long x) {    if (!is_neg) {        if (integer + x >= 0) {            integer += x;        } else {            bool last = false;            for (int i = len - 1; i >= 0; i--) {                if (last || data[i]) {                    data[i] = mo - data[i] - last;                    last = true;                } else {                    last = false;                }            }            integer = -x - integer - last;            is_neg = true;        }    } else {        if (integer - x >= 0) {            integer -= x;        } else {            bool last = false;            for (int i = len - 1; i >= 0; i--) {                if (last || data[i]) {                    data[i] = mo - data[i] - last;                    last = true;                } else {                    last = false;                }            }            integer = x - integer - last;            is_neg = false;        }    }    return *this;}Decimal & Decimal::operator += (int x) {    return *this += (long long)x;}Decimal & Decimal::operator -= (int x) {    return *this += (long long)-x;}Decimal & Decimal::operator -= (long long x) {    return *this += -x;}Decimal & Decimal::operator /= (int x) {    if (x < 0) {        is_neg ^= 1;        x = -x;    }    int last = integer % x;    integer /= x;    for (int i = 0; i < len; i++) {        long long tmp = 1LL * last * mo + data[i];        data[i] = tmp / x;        last = tmp - 1LL * data[i] * x;    }    if (is_neg && integer == 0) {        int i;        for (i = 0; i < len; i++) {            if (data[i] != 0) {                break;            }        }        if (i == len) {            is_neg = false;        }    }    return *this;}Decimal & Decimal::operator *= (int x) {    if (x < 0) {        is_neg ^= 1;        x = -x;    } else if (x == 0) {        init_zero();        return *this;    }    int last = 0;    for (int i = len - 1; i >= 0; i--) {        long long tmp = 1LL * data[i] * x + last;        last = tmp / mo;        data[i] = tmp - 1LL * last * mo;    }    integer = integer * x + last;    return *this;}Decimal operator - (const Decimal &a) {    Decimal ret = a;    // -0 = 0    if (!ret.is_neg && ret.integer == 0) {        int i;        for (i = 0; i < Decimal::len; i++) {            if (ret.data[i] != 0) break;        }        if (i < Decimal::len) {            ret.is_neg = true;        }    } else {        ret.is_neg ^= 1;    }    return ret;}Decimal operator + (const Decimal &a, int x) {    Decimal ret = a;    return ret += x;}Decimal operator + (int x, const Decimal &a) {    Decimal ret = a;    return ret += x;}Decimal operator + (const Decimal &a, long long x) {    Decimal ret = a;    return ret += x;}Decimal operator + (long long x, const Decimal &a) {    Decimal ret = a;    return ret += x;}Decimal operator - (const Decimal &a, int x) {    Decimal ret = a;    return ret -= x;}Decimal operator - (int x, const Decimal &a) {    return -(a - x);}Decimal operator - (const Decimal &a, long long x) {    Decimal ret = a;    return ret -= x;}Decimal operator - (long long x, const Decimal &a) {    return -(a - x);}Decimal operator * (const Decimal &a, int x) {    Decimal ret = a;    return ret *= x;}Decimal operator * (int x, const Decimal &a) {    Decimal ret = a;    return ret *= x;}Decimal operator / (const Decimal &a, int x) {    Decimal ret = a;    return ret /= x;}Decimal operator + (const Decimal &a, const Decimal &b) {    if (a.is_neg == b.is_neg) {        Decimal ret = a;        bool last = false;        for (int i = Decimal::len - 1; i >= 0; i--) {            ret.data[i] += b.data[i] + last;            if (ret.data[i] >= Decimal::mo) {                ret.data[i] -= Decimal::mo;                last = true;            } else {                last = false;            }        }        ret.integer += b.integer + last;        return ret;    } else if (!a.is_neg) {        // a - |b|        return a - -b;    } else {        // b - |a|        return b - -a;    }}Decimal operator - (const Decimal &a, const Decimal &b) {    if (!a.is_neg && !b.is_neg) {        if (a >= b) {            Decimal ret = a;            bool last = false;            for (int i = Decimal::len - 1; i >= 0; i--) {                ret.data[i] -= b.data[i] + last;                if (ret.data[i] < 0) {                    ret.data[i] += Decimal::mo;                    last = true;                } else {                    last = false;                }            }            ret.integer -= b.integer + last;            return ret;        } else {            Decimal ret = b;            bool last = false;            for (int i = Decimal::len - 1; i >= 0; i--) {                ret.data[i] -= a.data[i] + last;                if (ret.data[i] < 0) {                    ret.data[i] += Decimal::mo;                    last = true;                } else {                    last = false;                }            }            ret.integer -= a.integer + last;            ret.is_neg = true;            return ret;        }    } else if (a.is_neg && b.is_neg) {        // a - b = (-b) - (-a)        return -b - -a;    } else if (a.is_neg) {        // -|a| - b        return -(-a + b);    } else {        // a - -|b|        return a + -b;    }}Decimal operator + (const Decimal &a, double x) {    return a + Decimal(x);}Decimal operator + (double x, const Decimal &a) {    return Decimal(x) + a;}Decimal operator - (const Decimal &a, double x) {    return a - Decimal(x);}Decimal operator - (double x, const Decimal &a) {    return Decimal(x) - a;}Decimal & Decimal::operator += (double x) {    *this = *this + Decimal(x);    return *this;}Decimal & Decimal::operator -= (double x) {    *this = *this - Decimal(x);    return *this;}Decimal & Decimal::operator += (const Decimal &b) {    *this = *this + b;    return *this;}Decimal & Decimal::operator -= (const Decimal &b) {    *this = *this - b;    return *this;}// ---------- decimal lib end ----------#include <bits/stdc++.h>#define gc getchar()#define N 8009#define K(a,b) double(y[a]-y[b])/((a)-(b))using namespace std;int n,k,p,W,a[N],b[N],m,g[15][N],q[N],ed[15],h[N];double f[15][N],y[N];int read(){    char ch;    int x=1;    while (ch=gc,ch<'0'||ch>'9') if (ch=='-') x=-1;    int s=ch-'0';    while (ch=gc,ch>='0'&&ch<='9') s=s*10+ch-'0';    return s*x;}//most 14 kindsint main(){    n=read(),k=read(),p=read();    for (int i=1;i<=n;i++)    {        h[i]=read();        if (h[i]<h[1]) n--,i--;    }    sort(h+1,h+n+1);    for (int i=2;i<=n;i++) h[i]+=h[i-1];    for (int i=1;i<=n;i++) f[0][i]=h[1];    k=min(k,n-1);    W=min(k,14);    for (int i=1;i<=W;i++)    {        int l=1,r=0;        for (int j=2;j<=n;j++)        {            y[j-1]=h[j-1]-f[i-1][j-1];            while (l<r&&K(q[r-1],q[r])>=K(q[r],j-1)) r--;            q[++r]=j-1;            y[j+1]=h[j];            while (l<r&&K(q[l],j+1)<=K(q[l+1],j+1)) l++;            f[i][j]=K(q[l],j+1);            g[i][j]=q[l];        }    }    ed[W]=n-(k-W);    for (int i=W;i;i--) ed[i-1]=g[i][ed[i]];    Decimal ans=Decimal(h[1]);    for (int i=1;i<=W;i++)        ans=(ans+h[ed[i]]-h[ed[i-1]])/(ed[i]-ed[i-1]+1);    for (int i=ed[W]+1;i<=n;i++)        ans=(ans+h[i]-h[i-1])/2;    cout<<ans.to_string(p+2)<<endl;    return 0;}

D2T3：
似乎是一个提答题。。（smg，第一次遇见哎）？
前3个点好像是送分的？
第5个点我好像也会。。
剩下似乎都只会骗分啦。
似乎很复杂。。。还是丢链接跑。。

https://wenku.baidu.com/view/cc339a05551810a6f4248628.html

6、7、9、10的程序：

#include <bits/stdc++.h>using namespace std;int cnt;int in(){    puts("I");    return ++cnt;}void out(int x){    printf("O %d\n",x);    ++cnt;}int add(int i,int j){    printf("+ %d %d\n",i,j);    return ++cnt;}int C(int i,string j){    printf("C %d %s\n",i,j.c_str());    return ++cnt;}int rev(int i){    printf("- %d\n",i);    return ++cnt;}int L(int i,int j){    printf("< %d %d\n",i,j);    return ++cnt;}int R(int i,int j){    printf("> %d %d\n",i,j);    return ++cnt;}int S(int i){    printf("S %d\n",i);    return ++cnt;}string zeros(int n){    string s;    for (int i=1;i<=n;i++) s+='0';    return s;}void x_to_bits(int now,int *bits){    now=L(now,500);    long long t=702955280397374434ll;    char now_string[100];    for (int i=31;i;i--)    {        sprintf(now_string,"-%lld",t);        bits[i]=S(C(now,now_string+zeros(142)));        now=add(now,rev(L(bits[i],500+i)));        t/=2;    }    bits[0]=R(now,500);}int min_and_0(int x){    int p=L(S(L(C(x,"0."+zeros(29)+"1"),500)),151);    int y=S(add(R(x,150),p));    return add(L(C(y,"-0.5"),152),rev(p));}void solve6(){    cnt=0;    int bits[32],now=in();    x_to_bits(now,bits);    for (int i=31;i>=0;i--) out(bits[i]);}void solve7(){    cnt=0;    int a=in(),b=in(),bits[3][32];    x_to_bits(a,bits[0]);    x_to_bits(b,bits[1]);    for (int i=0;i<=31;i++)    {        int now=add(bits[0][i],bits[1][i]);        bits[2][i]=add(now,rev(L(S(L(C(now,"-1.5"),500)),1)));    }    for (int i=1;i<=31;i++) bits[2][i]=L(bits[2][i],i);    for (int i=1;i<=31;i++) bits[2][0]=add(bits[2][0],bits[2][i]);    out(bits[2][0]);}void solve9(){    cnt=0;    int a[17];    for (int i=1;i<=16;i++) a[i]=in();    for (int i=1;i<=16;i++)        for (int j=i+1;j<=16;j++)        {            int s=add(a[i],a[j]);            a[i]=add(a[i],min_and_0(add(rev(a[i]),a[j])));            a[j]=add(s,rev(a[i]));        }    for (int i=1;i<=16;i++) out(a[i]);}int p(int x){    return S(L(x,600));}int get(int x,int y){    int p=L(x,151);    return add(L(C(S(add(rev(p),R(y,150))),"-0.5"),152),p);}void solve10(){    cnt=0;    int a=in(),b=in(),m=in(),m_fu=rev(m);    int mm=C(m,"-0."+zeros(29)+"1");    int bits[32],Bits[32];    int zero=R(m,1000),ans;    x_to_bits(a,Bits);    x_to_bits(b,bits);    int A=Bits[31];    A=add(A,get(p(add(mm,rev(A))),m_fu));    for (int i=30;i>=0;i--)    {        A=add(add(A,A),Bits[i]);        A=add(A,get(p(add(mm,rev(A))),m_fu));    }    a=A;    ans=get(C(rev(bits[0]),"1"),a);    ans=add(ans,get(p(add(mm,rev(ans))),m_fu));    for (int i=1;i<32;i++)    {        a=add(a,a);        a=add(a,get(p(add(mm,rev(a))),m_fu));        ans=add(ans,get(C(rev(bits[i]),"1"),a));        ans=add(ans,get(p(add(mm,rev(ans))),m_fu));    }    out(ans);}int main(){    freopen("nodes6.out","w",stdout);    solve6();    freopen("nodes7.out","w",stdout);    solve7();    freopen("nodes9.out","w",stdout);    solve9();    freopen("nodes10.out","w",stdout);    solve10();    return 0;}