Loi队内胡策 round 1 - 5

之前进行了一星期的胡策，一直坑着没写233，今天决定补上。
由于不可描述的原因，进行了为期一周的队内胡策(互相伤害)，每次出题人不同，题目方向不同，但所有算法确实是在noip范围内，都是套路啊，被学弟学妹们虐的死去活来233。

round 1

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round 1 由swc , imcy , xczhw三位老年选手出题╮(╯▽╰)╭。题目大都可以找到原题233.

T1

swc正在找题的时候，imcy突然发现这个题只要换下输入就非常劲233，于是就有了t1，有了这个样例

实际效果非常有视觉冲击力233.
所以说题解就是求个连通块。
std如下

#include<iostream>#include<cstdio>#include<queue>using namespace std;int map[500][500];struct doubi{    int x,y;};int n,m;queue<doubi> q;int x_x[13]={0,0,0,1,-1,1,1,-1,-1,0,0,2,-2};int y_y[13]={0,1,-1,0,0,1,-1,1,-1,2,-2,0,0};bool can(int x,int y){    if(x>=1&&x<=n&&y>=1&&y<=m)        return true;    return false;}void bfs(int i,int j){    doubi s;    s.x=i,s.y=j;    q.push(s);    while(!q.empty())    {        doubi u=q.front();        q.pop();        int x=u.x,y=u.y;        for(int i=1;i<=12;i++){            int X=x+x_x[i];            int Y=y+y_y[i];            if(can(X,Y)&&map[X][Y]){                doubi nxt;                nxt.x=X,nxt.y=Y;                q.push(nxt);                map[X][Y]=0;            }        }    }}int main(){    freopen("chinese.in","r",stdin);    freopen("chinese.out","w",stdout);        scanf("%d%d",&n,&m);        for(int i=1;i<=n;i++){            for(int j=1;j<=m;j++){                string x;                cin>>x;                if(x=="井号"){                    map[i][j]=1;                }            }        }        int ans=0;        for(int i=1;i<=n;i++){            for(int j=1;j<=m;j++){                if(map[i][j]){                    bfs(i,j);                    ans++;                }            }        }        cout<<ans<<endl;    fclose(stdin);    fclose(stdout);    return 0;}

T2

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一个USACO的题目，当时做的时候一眼建图，然后打了个最短路233，发现样例过不了，仔细阅读题目后，神tm最短路，把每个点都连起来再加上几个井，明明是最小生成树233，题目沿用了luogu.org上的翻译.
std如下

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>using namespace std;const int size = 200010;int n;int num[size];struct dc{    int f,t,d;}l[size];int tot = 1;void build(int f,int t,int d){    l[tot].f = f;    l[tot].t = t;    l[tot].d = d;    tot ++;}int f[size];int find(int x){    if(f[x] == x)        return x;    return f[x] = find(f[x]);}bool cmp(dc a,dc b){    return a.d < b.d;}void init(){    tot = 1;    memset(num,0,sizeof(num));    memset(l,0,sizeof(l));    memset(f,0,sizeof(f));}int main(){    freopen("farmer_m.in","r",stdin);    freopen("farmer_m.out","w",stdout);    scanf("%d",&n);    for(int i = 1 ; i <= n ; i ++)        scanf("%d",&num[i]) , f[i] = i;    for(int i = 1 ; i <= n ; i ++)        for(int j = 1 ; j <= n ; j ++)        {            int d;            scanf("%d",&d);            build(i,j,d);        }    for(int i = 1 ; i <= n ; i ++)        build(0,i,num[i]);    sort(l+1,l+tot,cmp);    int ans = 0;    for(int i = 1 ; i < tot ; i ++)    {        int ff = l[i].f , ft = l[i].t;        if(find(ff) != find(ft))        {            f[find(ff)] = find(ft);            ans += l[i].d;        }    }    printf("%d\n",ans);    fclose(stdin);    fclose(stdout);    return 0;}

T3

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题目是faebdc神犇曾给提供的模拟赛的t3，题面被魔改的非常辣眼233.一个树上LCA乱搞的好题，具体见代码
std如下

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#define KILL puts("haha");using namespace std;typedef long long ll;const ll MAXN = 100000 + 5;const ll MAXM = 100000 + 5;const ll P = 1e9 + 7;struct edge{    ll f,t,v;}l[MAXM << 1];ll first[MAXN],next[MAXM << 1],tot;ll n;void init(){    memset(first,0xfff,sizeof(first));    tot = 0;    return;}void build(ll f,ll t,ll v){    l[++tot] = (edge){f,t,v};    next[tot] = first[f];    first[f] = tot;    return;}ll fa[MAXN][35],deep[MAXN];void get_fa(){    for(int j = 1;j <= 30;j ++)        for(int i = 1;i <= n;i ++)            fa[i][j] = fa[fa[i][j - 1]][j - 1];}ll lca(ll x,ll y){    if(deep[x] < deep[y])        swap(x,y);    for(int j = 30;j >= 0;j --)        if(deep[fa[x][j]] >= deep[y])            x = fa[x][j];    if(x == y)        return x;    for(int j = 30;j >= 0;j --)        if(fa[x][j] != fa[y][j])            x = fa[x][j],y = fa[y][j];    return fa[x][0];}ll sz[MAXN];void dfs(ll x,ll f){    fa[x][0] = f;    sz[x] = 1;    deep[x] = deep[f] + 1;    for(int i = first[x];~i;i = next[i])        if(l[i].t != f)        {            dfs(l[i].t,x);            sz[x] += sz[l[i].t];        }    return;}ll root[MAXN];ll ci(ll x,ll y){    if(deep[x] < deep[y])        swap(x,y);    return sz[x] * (n - sz[x]);}void dfs(ll x){    root[x] = root[fa[x][0]] + ci(fa[x][0],x);    for(int i = first[x];~i;i = next[i])        if(l[i].t != fa[x][0])            dfs(l[i].t);    return;}ll ans = 0;void solve(){    for(int i = 1;i <= tot;i += 2)        ans += (ci(l[i].f,l[i].t) % P * l[i].v % P) % P,ans %= P;    ans %= P;    return;}ll m;ll f,t,v;ll read(){    ll x = 0 , f = 1;    char in = getchar();    while(in < '0' || in > '9')    {        if(in == '-')            f = -1;        in = getchar();    }    while(in >= '0' && in <= '9')    {        x = (x << 3) + (x << 1) + in - '0';        in = getchar();    }    return x * f;}int main(){    freopen("MaHouShaoJiuDQS.in","r",stdin);    freopen("MaHouShaoJiuDQS.out","w",stdout);    init();    n = read();    for(int i = 1;i < n;i ++)    {        f = read() , t = read() , v = read();        while(v < 0)            v += P;        v %= P;        build(f,t,v);        build(t,f,v);    }    dfs(1,0);    get_fa();    dfs(1);    solve();    printf("%lld\n",ans);    m = read();    for(int i = 1;i <= m;i ++)    {        f = read() , t = read() , v = read();        while(v < 0)            v += P;        v %= P;        ans = ((ans % P + (root[f] % P + root[t] % P - ((root[lca(f,t)] % P) << 1) % P) % P * (v % P)) % P) % P;        while(ans < 0)            ans += P;        printf("%lld\n",ans % P);        ans %= P;    }    fclose(stdin);    fclose(stdout);    return 0;}

这次胡策学弟(妹)们表现的还不错，该A的A，该暴力的暴力233.

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round 2

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round 2 由 black，francis，knight，qer出题。

T1

看完题一脸蒙蔽233，开始手推公式，半小时后放弃，一个半小时后打出表来发现正解233，知道正解后才想起来之前好像听哪位学长说过fib的公比为(√5 - 1)/2，于是A掉T1，差点被套路到233。
代码如下

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;unsigned long long f[450];int main(){    freopen("DQSjangke.in","r",stdin);    freopen("DQSjangke.out","w",stdout);    unsigned long long n;    int m;    cin>>n;    f[0] = 1;    f[1] = 1;    for(int i = 2 ;  ; i ++)    {        f[i] = f[i-1] + f[i-2];        if(f[i] > n || f[i] < 0)        {            m = i;            break;        }    }    m --;    cout<<f[m-1]<<"/"<<f[m];    fclose(stdin);    fclose(stdout);    return 0;}

T2

看完题还是一脸蒙蔽，T1已经用了一半时间，这个题也没多想，感觉贪心好像可以，每次拿一个尽量大且合法的分离器装上，然后简单想了下证明，复杂度O(k)，就直接T3了，考完才发现k是10^9,只拿了部分分+手动读入骗来的分，加个二分就可以过了，还可以O(1)求解，然而我不会233。
std如下

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;typedef long long ll;void read(ll &n){    n = 0;    char a = getchar();    while(a<'0'||a>'9')        a = getchar();    while(a>='0'&& a<='9')    {        n*=10;        n+=a-'0';        a = getchar();    }}int main(){    freopen("project.in", "r", stdin);    freopen("project.out", "w", stdout);    ll n, k, d, t;    read(n);read(k);    d=k*k - k - 2*n + 2;    if(d < 0)        puts("-1");    else    {        t = (int)(sqrt(d));        printf("%lld", k-1-t+(t*t+t>d));    }}

加了一点点特技的背包,然而当时没时间了比较慌，没想到n^2算法,于是大力暴力n^3,神tmA了233.
代码如下

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#define ll long longusing namespace std;ll read(){    int x = 0 , f = 1;    char in = getchar();    while(in < '0' || in > '9')    {        if(in == '-')            f = -1;        in = getchar();    }    while(in >= '0' && in <= '9')    {        x = x * 10 + in - '0';        in = getchar();    }    return x * f;}ll f[4010];int m,n;ll w[2010],a[2010],b[2010];int main(){    freopen("candy.in","r",stdin);    freopen("candy.out","w",stdout);    scanf("%d%d",&m,&n);    for(int i = 1 ; i <= n ; i ++)        w[i] = read() , a[i] = read() , b[i] = read();    for(int i = 1 ; i <= n ; i ++)        for(int j = m ; j >= w[i] ; j --)            for(int k = 1 ; j - k * w[i] >= 0 ; k ++)                f[j] = max(f[j],f[j-k*w[i]]+k*(a[i]+b[i])-(k-1)*b[i]);    printf("%lld\n",f[m]);    fclose(stdin);    fclose(stdout);    return 0;}/*100 210 2 120 1 5068*/

T4

题解强行noip233，就是把数位dp的过程搞了下，数位dp不可免233，非noip内容这里就不bb了。
std如下

#include <iostream>#include <cstdio>#include <cstring>#include <queue>using namespace std;typedef long long LL;const int MAXN = 65;LL c[MAXN][MAXN];LL bin[MAXN];void read(LL &n){    n = 0;    char a = getchar();    while(a<'0'||a>'9')        a = getchar();    while(a>='0'&& a<='9')    {        n*=10;        n+=a-'0';        a = getchar();    }}void init(){    for(int i = 0; i < MAXN; i ++)        c[i][0] = 1;    for(int i = 1; i < MAXN; i ++)        for(int j = 1; j < MAXN; j ++)            c[i][j] = c[i-1][j] + c[i-1][j-1];}LL Get_bin(LL x){    bin[0] = 0;    while(x)    {        bin[++bin[0]] = x%2;        x >>= 1;    }}LL Get_Rn(LL x){    LL sum = 0;    Get_bin(x); // 获得二进制数    // 长度小于 x    for(int i = bin[0]-1; i > 1; i --)        for(int j = (i+1)/2; j <= i-1; j ++)            sum += c[i-1][j];//  sum = 0;    // 长度等于 x // 需要的 0 恒大于 (bin[0]+1)/2    LL zero = 0;    for(LL i = bin[0]-1; i >= 1; i --) // 本来就有一个 1        if(bin[i])            for(LL j = ((bin[0]+1)/2)-1-zero; j <= i-1; j ++)                sum += c[i-1][j];        else zero ++;    return sum;}int main(){    freopen("Oierdqs.in","r",stdin);    freopen("Oierdqs.out","w",stdout);    LL a, b;    read(a);read(b);    init();//  cout << Get_Rn(b+1) << endl;    printf("%llu", Get_Rn(b+1)-Get_Rn(a));    return 0;}

round 2的t1，t2都挺有意思的，不失为一种思路，数位dp我刚接触没几天，对t4不做评价，t3有点裸了，恩恩，不过学弟们的题贼带劲。

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round 3

round 3 是 jhz lcy ljx msy zzk出题，题目贼劲啊，神tmT2就到ctsc了233.

T1

一个有坑的暴力题，注意下边界没啥问题。
代码如下

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#define ll long longusing namespace std;ll num[511][511];int n;int px,py;bool check(){    ll ans = 0 , temp = 0;    for(int i = 1 ; i <= n ; i ++)        ans += num[i][i];    for(int i = 1 ; i <= n ; i ++)        temp += num[i][n-i+1];    if(ans != temp)        return false;    for(int i = 1 ; i <= n ; i ++)    {        temp = 0;        for(int j = 1 ; j <= n ; j ++)            temp += num[i][j];        if(temp != ans)            return false;    }    for(int i = 1 ; i <= n ; i ++)    {        temp = 0;        for(int j = 1 ; j <= n ; j ++)            temp += num[j][i];        if(temp != ans)            return false;    }    return true;}int main(){    freopen("business.in","r",stdin);    freopen("business.out","w",stdout);    scanf("%d",&n);    for(int i = 1 ; i <= n ; i ++)        for(int j = 1 ; j <= n ; j ++)        {            scanf("%lld",&num[i][j]);            if(num[i][j] == 0)                px = i , py = j;        }    int pos = 0;    if(n == 1)    {        puts("1");        fclose(stdin);        fclose(stdout);        return 0;    }    else if(px == 1)        pos = 2;    else if(px == n)        pos = n - 1;    else        pos = px + 1;    ll ans = 0;    for(int i = 1 ; i <= n ; i ++)        ans += num[pos][i];    for(int i = 1 ; i <= n ; i ++)        ans -= num[px][i];    num[px][py] = ans;    if(check())        printf("%lld\n",ans);    else        puts("-1");    fclose(stdin);    fclose(stdout);    return 0;}

T2

对字符串的题一直不敏感233，做到这一脸懵逼，思考过后决定打暴力233.
考完后知道原题是ctsc的，吓尿了233，正解是hash。
std如下

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <queue>using namespace std;typedef unsigned long long ULL; const int p = 2333;char s[30010][550];ULL hash[50000], py[500], ls[50000];int main(){    freopen("gbwl.in", "r", stdin);    freopen("gbwl.out", "w", stdout);    int len, n, si;    scanf("%d%d%d", &n, &len, &si);    py[0] = 1;    for(int i = 1; i <= 210; i++)        py[i] = py[i - 1] * p;    for(int i = 1; i <= n; i++)        scanf("%s", s[i] + 1);    for(int i = 1; i <= n; i++)        for(int j = 1; j <= len; j++)            hash[i] = hash[i] * p + s[i][j];    int pr = 1, ans = 0;    for(int i = 1; i <= len; i++)    {        for(int j = 1; j <= n; j++)            ls[j] = hash[j] - s[j][i] * py[len - i];        sort(ls + 1, ls + n + 1);        for(int j = 2; j <= n; j++)        {            if(ls[j] == ls[j - 1])            {                ans += pr;                pr ++;            }            else                pr = 1;        }    }    printf("%d", ans);    return 0;

T3

去年noiptg组day2t3原题，然而我现在还是不会233，直接上分段函数，palapala打了200+行，评测的时候发现好像打次了！？跪惨。
std如下

#include <iostream>#include <cstdio>#include <algorithm>using namespace std;const int SZ = 300010;struct Plan{    int u, v, lca, dis;}p[SZ];struct Edge{    int f, t, d;}es[SZ << 1];int first[SZ << 1], nxt[SZ << 1], tot = 1, cnt; int n, m, jump[SZ][30], dis[SZ], in[SZ];int recover[SZ], cost[SZ], depth[SZ];   int u, v, w, TOT = 0, l, r, mid;bool vis[SZ];int read(){    int f = 1, num = 0;    char c = getchar();    while(c < '0' || '9' < c)    {        if(c == '-') f = -1;         c = getchar();    }    while('0' <= c && c <= '9')    {        num = num * 10 + (c - '0');        c = getchar();    }    return num *= f;}inline void build(int ff, int tt, int dd){    es[++tot] = (Edge){ff, tt, dd};    nxt[tot] = first[ff];    first[ff] = tot;    es[++tot] = (Edge){tt, ff, dd};    nxt[tot] = first[tt];    first[tt] = tot;}inline void dfs(int u){    for(int i = 1; i <= 17; ++i)        jump[u][i] = jump[jump[u][i - 1]][i - 1];    vis[u] = 1;    in[++cnt] = u;    for(int i = first[u]; i; i = nxt[i])    {        int v = es[i].t;        if(vis[v]) continue;        jump[v][0] = u;        depth[v] = depth[u] + 1;        dis[v] = dis[u] + es[i].d;        cost[v] = es[i].d;        dfs(v);     }}int get_lca(int uu, int vv){    if(depth[uu] < depth[vv]) swap(uu, vv);    int t = depth[uu] - depth[vv];    for(int i = 0; i <= 17; ++i)        if((t >> i) & 1) uu = jump[uu][i];    for(int i = 17; i >= 0; --i)        if(jump[uu][i] != jump[vv][i])            uu = jump[uu][i], vv = jump[vv][i];    if(uu == vv) return uu;    return jump[uu][0];}bool check(int mid){    for(int i = 1; i <= n; ++i) recover[i] = 0;    int tot = 0, maxn = 0;    for(int i = 1; i <= m; ++i)        if(p[i].dis > mid)        {            ++tot;            maxn = max(p[i].dis, maxn);            ++recover[p[i].u]; ++recover[p[i].v]; recover[p[i].lca]-=2;        }        //else break;    if(!tot) return true;    for(int i = cnt; i > 1; --i) recover[jump[in[i]][0]] += recover[in[i]];    for(int i = 2; i <= n; ++i)        if(maxn - cost[i] <= mid && recover[i] == tot) return true;    return false;}int main(){    int __size__ = 1024 << 15;    char * __p__ = (char *) malloc(__size__) + __size__;    __asm__("movl %0, %%esp\n" :: "r"(__p__));    freopen("foolish.in", "r", stdin);    freopen("foolish.out", "w", stdout);    n = read(); m = read();    for(int i = 1; i < n; ++i)    {        u = read(), v = read(), w = read();        build(u, v, w); TOT += w;    }    depth[1] = 1;    dfs(1);    for(int i = 1; i <= m; ++i)    {        p[i].u = read(); p[i].v = read();        //cout<<depth[p[i].u]<<" "<<depth[p[i].v]<<endl;        p[i].lca = get_lca(p[i].u, p[i].v);        p[i].dis = dis[p[i].u] + dis[p[i].v] - (dis[p[i].lca] << 1);        //cout<<p[i].lca<<endl;    }    l = 0, r = TOT;    while(l <= r)    {        mid = (l + r) >> 1;        if(check(mid)) r = mid - 1;        else l = mid + 1;    }    printf("%d", l);    fclose(stdin);    fclose(stdout);    return 0;}

胡策以来跪的最惨的一次考试233，基本都是暴力分。

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round 4

出题 by 天线 DL fluency a 帝江

考试时的cy:T1裸中余定，然而我不会，上暴力233，T2一模拟！？，瞎搞一搞，好像会有精度问题，T3sx题直接过，T4什么玩意，好像线段树能搞？拿线段树搞一搞。于是就有了T1暴力，T2翻车，T3AC，T4暴力的cy ╮(╯▽╰)╭。

T1

中国剩余定理Orz，不过还是暴力大法好233.
std如下

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long LL;const int MAXN=1000+10;int a[MAXN],b[MAXN];LL exgcd(LL a,LL b,LL &x,LL &y)//扩展欧几里得 {    if(b==0)    {        x=1;        y=0;        return a;    }    else    {        LL p=exgcd(b,a%b,x,y);        LL t=x;        x=y;        y=t-a/b*x;        return p;    }}/*x%a[i]==b[i]*/LL China(int n)//中国剩余定理 {    LL M=1,d,x=0,y;    for(int i=1;i<=n;i++)        M*=(LL)a[i];    for(int i=1;i<=n;i++)    {        LL w=M/(LL)a[i];        exgcd(w,a[i],d,y);        x=(x+d*w*b[i])%M;      }    return (x+M)%M;}int main(){    freopen("Question.in","r",stdin);    freopen("Question.ans","w",stdout);    int n;    scanf("%d",&n);    for(int i=1;i<=n;i++)        scanf("%d%d",&a[i],&b[i]);    printf("%lld\n",China(n));    return 0;}

T2

确实就是一个模拟，不过我好像搞错了什么233，直接上代码
std如下

#include<cstdio>#include<cmath>#include<algorithm>#define inf 598460606#define LL long longusing namespace std;int n,m,per=1;int lt,ld,nt=1;char ch[5];double t[210000],d[210000],x,nowt;int main(){    freopen("dig.in","r",stdin);    freopen("dig.out","w",stdout);    scanf("%d%d",&n,&m);    for (int i=1;i<=n;i++)     {        scanf("%s%lf",ch,&x);        if (ch[0]=='T')t[++lt]=x;        else d[++ld]=x;     }    d[++ld]=0.0;    d[++ld]=m;    sort(d+1,d+ld+1);    sort(t+1,t+lt+1);    for(int i=1;i<ld;i++)      {        double nd=d[i];        while (nd<d[i+1]&&nt<=lt&&nowt+(d[i+1]-nd)*per>t[nt])        {            nd+=(t[nt]-nowt)/per;            per++;            nowt=t[nt++];        }        nowt+=(d[i+1]-nd)*per;        per++;      }    printf("%I64d\n",(long long)nowt);}

T3

出题人好像是想要考代码细节处理？好多人都在这题上翻车了233，注意下并不只是判断最短路上是否存在负环就好吧。
代码如下

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#define ll long longusing namespace std;const int size = 2001000;int read(){    int x = 0 , f = 1;    char in = getchar();    while(in < '0' || in > '9')    {        if(in == '-')            f = -1;        in = getchar();    }    while(in >= '0' && in <= '9')    {        x = x * 10 + in - '0';        in = getchar();    }    return x * f;}int head[size],nxt[size];ll dist[size];int tot = 1;struct gtnd{    int t;    ll d;}l[size];int n,m;void build(int f,int t,int d){    l[tot].t = t;    l[tot].d = d;    nxt[tot] = head[f];    head[f] = tot ++;}queue < int > q;bool use[size];bool vis[size];int tim[size];void dfs(int u){    vis[u] = 1;    for(int i = head[u] ; i ; i = nxt[i])        if(!vis[l[i].t])            dfs(l[i].t);}void init(){    while(!q.empty())        q.pop();    for(int i = 1 ; i <= n ; i ++)        dist[i] = 214748364111ll;    for(int i = 1 ; i <= n ; i ++)        tim[i] = 0;}bool spfa(int s){    init();    dist[s] = 0;    use[s] = 1;    q.push(s);    while(!q.empty())    {        int f = q.front();        q.pop();        use[f] = 0;        for(int i = head[f] ; i ; i = nxt[i])        {            int t = l[i].t;            if(dist[t] > dist[f] + l[i].d)            {                dist[t] = dist[f] + l[i].d;                tim[t] = tim[f] + 1;                if(tim[t] > 2 * n)                    return false;                if(!use[t])                {                    use[t] = 1;                    q.push(t);                }            }        }    }    return true;}int main(){    freopen("EasySSSP.in","r",stdin);    freopen("EasySSSP.out","w",stdout);    n = read() , m = read();    int s = read();    for(int i = 1 ; i <= m ; i ++)    {        int f = read() , t = read() , d = read();        build(f,t,d);    }    for(int i = 1 ; i <= n ; i ++)    {        if(!vis[i])        {            dfs(i);            if(!spfa(i))            {                puts("-1");                fclose(stdin);                fclose(stdout);                return 0;            }        }    }    spfa(s);    for(int i = 1 ; i <= n ; i ++)    {        if(dist[i] == 214748364111ll)            puts("NoPath");        else            printf("%lld\n",dist[i]);    }    fclose(stdin);    fclose(stdout);     return 0;}/*5 5 11 2 12 1 13 4 24 5 -25 3 2*/

T4

红太阳的题太强啦，只会暴力呀，乱搞大失败呀233，大体意思是分下块然后用set维护下？
std如下

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int f[300005],num[300005];int mn[605],ans[300005];int n;char opt[100005];int S[1000005],top;int find(int i){    while(f[i]!=i)    {        S[++top]=i;        i=f[i];    }    while(top)    {        f[S[top]]=i;        top--;    }    return i;}int cnt[300005];char s[5];int main(){    freopen("Math.in","r",stdin);    freopen("Math.out","w",stdout);    scanf("%d",&n);    for (int i=1;i<=300000;i++) f[i]=i+1;    memset(mn,127,sizeof(mn));    for (int i=1;i<=n;i++)    {        scanf("%s",s);        opt[i]=s[0];        scanf("%d",&num[i]);        if (opt[i]=='A')        {            for (int j=1;j<=540;j++) mn[j]=min(mn[j],num[i]%j);            f[num[i]]=num[i];cnt[num[i]]++;        }        else if (num[i]<=540) ans[i]=mn[num[i]];    }    for (int i=n;i;i--)    {        if (opt[i]=='A')        {            cnt[num[i]]--;            if(cnt[num[i]]==0)                f[num[i]]=f[num[i]+1];        }        else if(num[i]>540)        {            ans[i]=find(1)%num[i];            for (int j=num[i];j<=300000;j+=num[i])            {                int x=find(j);                if (x) ans[i]=min(ans[i],x%num[i]);            }        }    }    for (int i=1;i<=n;i++)         if (opt[i]=='B') printf("%d\n",ans[i]);    return 0;}

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round 5

出题 by ChlorineHikari，Yuan，Meico，Summer， （题目顺序）Peacefuldoge（后期）
最后的一次胡策，题目偏向模板。
考试时的cy : T1不就模拟吗233，T2一sx压位搜索，wocT3K短路然而我没学过A*，wocT4一树形DP然而我还是没学过╮(╯▽╰)╭。

T1

模拟搞一搞，据说是cf上的原题。
std如下

//Codeforces Round #375 (Div. 2) B#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>using namespace std;const int maxn=20000;char hah[maxn];int l[maxn];int r[maxn];int n;int ans2=0;void haha(int ll,int rr){    for(int i=ll+1;i<rr;)    {        if(hah[i]=='_')        i++;        else        {            ans2++;            while(hah[i]!='_'&&i<rr)            {                i++;            }        }    }}int main(){    freopen("letter.in","r",stdin);    freopen("letter.osu","w",stdout);    int ans1=0;    scanf("%d",&n);    int lc=0;    int rc=0;    for(int i=1;i<=n;i++)    {        cin>>hah[i];        if(hah[i]=='(')        {            l[++lc]=i;        }        if(hah[i]==')')        {            r[++rc]=i;        }    }    for(int i=1;i<=n;)    {        if(hah[i]=='_')        {            i++;        }        else if(hah[i]=='(')        {            while(i<=n)            {                i++;                if(hah[i]==')')                break;            }        }        else         {            if(hah[i]==')')            i++;            int cnt=0;            while(hah[i]!='_'&&hah[i]!='('&&i<=n)            {                i++;                cnt++;            }            ans1=max(ans1,cnt);        }    }    for(int i=1;i<=lc;i++)    {        haha(l[i],r[i]);    }    cout<<ans1<<" ";    printf("%d",ans2);    return 0;}

T2

经典的状压搜索问题，没啥好注意的。
代码如下

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>using namespace std;const int sz = 1 << 11;const int size = 214;struct gtnd{    int now;    int step;};int repair[size];int destory[size];int n,m;queue < gtnd > q;bool check(int x){    for(int i = 1 ; i <= n ; i ++)        if(!((x>>i) & (1)))            return false;    return true;}bool use[1000010];/*void print(int x){    for(int i = 1 ; i <= n ; i ++)        printf("%d ",(x >> i) & (1));    puts("");}*/int bfs(){    gtnd s;    s.now = 0;    s.step = 0;    use[s.now] = 1;    q.push(s);    while(!q.empty())    {        gtnd f = q.front();        q.pop();        if(check(f.now))            return f.step;        for(int i = 1 ; i <= m ; i ++)        {            gtnd nxt;            nxt.now = f.now;            nxt.now |= repair[i];            nxt.now &= (~destory[i]);            if(use[nxt.now] == 0)            {                use[nxt.now] = 1;                nxt.step = f.step + 1;                q.push(nxt);            }        }    }    return -1;}int main(){    freopen("glgjssy.In","r",stdin);    freopen("glgjssy.0ut","w",stdout);    scanf("%d%d",&n,&m);    for(int i = 1 ; i <= m ; i ++)    {        for(int j = 1 ; j <= n ; j ++)        {            int ins;            scanf("%d",&ins);            if(ins == 1)                repair[i] |= 1 << j;            else if(ins == -1)                destory[i] |= 1 << j;        }    }    int ans = bfs();    if(ans == -1)        puts("GG");    else        printf("%d\n",ans);    fclose(stdin);    fclose(stdout);    return 0;}/*3 21 0 1-1 1 03 31 -1 -1-1 1 -1-1 -1 13 31 -1 00 1 -10 0 11 0 1 / 0 0 00 1 0 / 1 0 0*/

T3

由于我对dij还A*都不感冒，考试的时候就打了个spfa求次短路糊弄了下233。
std如下

//POJ 2449 k短路 #include<algorithm>#include<cstring>#include<cstdio>#include<queue>using namespace std;const int inf=0x7fffffff;const int maxn=100000+500;struct Edge{    int to;    int d;    int next;}edge[maxn],edge1[maxn];int head[maxn],head1[maxn],h[maxn];int n,m,ss,tt,k;struct heap{    int p;    int dis;    int h;    bool operator <(const heap &hah)const     {        return dis+h>hah.dis+hah.h;     }};priority_queue<heap>q;int tott;void add(int f,int t,int d){    edge[++tott].to=t;    edge[tott].d=d;    edge[tott].next=head[f];    head[f]=tott;}int tot1;void add1(int f,int t,int d){    edge1[++tot1].to=t;    edge1[tot1].d=d;    edge1[tot1].next=head1[f];    head1[f]=tot1;}queue<int>q1;bool vis[maxn];void spfa(){    for(int i=1;i<=n;i++)        h[i]=inf;    q1.push(tt);    h[tt]=0;    while(!q1.empty())    {        int x=q1.front();        vis[x]=0;        q1.pop();        for(int i=head1[x];i;i=edge1[i].next)        {            Edge e=edge1[i];            if(h[e.to]>h[x]+e.d)            {                h[e.to]=h[x]+e.d;                if(!vis[e.to])                {                    vis[e.to]=1;                     q1.push(e.to);                }            }        }     }}int ans=-1;void DJ(){    int cnt=0;    heap hah;    hah.p=ss;    hah.dis=0;    hah.h=h[ss];    q.push(hah);    while(!q.empty())    {        heap pre=q.top();        q.pop();        int x=pre.p;        if(x==tt)        {            cnt++;            if(cnt==k)            {                ans=pre.dis;                return;            }        }        for(int i=head[x];i;i=edge[i].next)        {            Edge e=edge[i];            q.push((heap){e.to,pre.dis+e.d,h[e.to]});        }    } }int main(){    freopen("chemistry.in","r",stdin);    freopen("chemistry.out","w",stdout);    scanf("%d%d",&n,&m);    for(int i=1;i<=m;i++)    {        int a,b,c;        scanf("%d%d%d",&a,&b,&c);        add(a,b,c);        add1(b,a,c);    }    scanf("%d%d%d",&ss,&tt,&k);    if(ss==tt)    k++;    spfa();    DJ();    printf("%d",ans);    return 0;}

T4

原题codevs选课，多叉转二叉后跑下dp。
代码如下

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>using namespace std;const int size = 1010;const int sz = 200010;int val[size];int dist[size];int r[size];int len,n,m;int head[sz],nxt[sz],l[sz];int tot = 1;bool use[size];int f[size][size]; // f[i][j] 第 i 个节点 void build(int f,int t){    l[tot] = t;    nxt[tot] = head[f];    head[f] = tot ++;}struct gtnd{    int l, r;}tree[size];queue < int > q;int read(){    int x = 0 , f = 1;    char in = getchar();    while(in < '0' || in > '9')    {        if(in == '-')            f = -1;        in = getchar();    }    while(in >= '0' && in <= '9')    {        x = (x << 3) + (x << 1) + in - '0';        in = getchar();    }    return x * f;}int dfs(int u,int pick){    if(!u)        return 0;    if(f[u][pick])        return f[u][pick];    for(int i = 0 ; i <= pick - 1 ; i ++)        f[u][pick] = max(f[u][pick],dfs(tree[u].l,i)+val[u]+dfs(tree[u].r,pick-i-1));    f[u][pick] = max(f[u][pick],dfs(tree[u].r,pick));    return f[u][pick];}int main(){    freopen("dreamerpeacefuldoge.in","r",stdin);    freopen("dreamerpeacefuldoge.out","w",stdout);    n = read() , m = read();    for(int i = 1 ; i <= n ; i ++)    {        int par = read();        val[i] = read();        if(tree[par].l)            tree[i].r = tree[par].l;        tree[par].l = i;    }    printf("%d\n",dfs(tree[0].l,m));    fclose(stdin);    fclose(stdout);    return 0;}

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总结下，虽然我不怎么会总结
看到学弟(妹)们越来越套路我还是很滋磁的233，大家noip加油吧233。
All Loiers rp ++;