ZF_20160702_Region2013长春

B题
模拟(1+/5)/2进制，以前写过。

//{using namespace std;//#pragma warning(disable:4996)//#pragma comment(linker,"/STACK:102400000,102400000")#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<set>#include<map>#include<vector>#include<cmath>#define ll         long long#define clr(x,y)   memset(x,y,sizeof(x))#define ww         while#define r1(i,a,b)  for(int i=a;i< b;i++)#define r2(i,a,b)  for(int i=a;i<=b;i++)#define r3(i,a,b)  for(int i=a;i> b;i--)#define r4(i,a,b)  for(int i=a;i>=b;i--)#define r5(i,a,n)  for(int i=1;i<=n;i++)sf1(a[i])#define r6(i,a,s,t)  for(int i=s;i<=t;i++)pf("%d%c",a[i],i-(t)?' ':'\n')#define ss(x)      sf("%s",x)#define sf1(x)     sf("%d",&x)#define sf2(x,y)   sf("%d%d",&x,&y)#define sf3(x,y,z) sf("%d%d%d",&x,&y,&z)#define pd(x,y)    printf("%d%c",x,(char)y)#define sf         scanf#define pf         printf#define pct(x)     __builtin_popcount(x)#define ctz(x)     __builtin_ctz(x)const int inf=0x3f3f3f3f;const ll MOD=1e9+7;const int o=200;struct STC{    int a[420],l,r;    STC(){clr(a,0);}    void E(){clr(a,0);a[o]=1;l=r=o;}    void out()    {        r4(i,r,o)printf("%d",a[i]);putchar('.');r4(i,o-1,l)printf("%d",a[i]);    }    void get(char c[])    {        int len=strlen(c);        int pos;        r1(i,0,len)if(c[i]=='.')pos=i;        r=pos+199;l=pos+201-len;        r1(i,0,pos)a[r-i]=c[i]-'0';        r1(i,pos+1,len)a[pos+200-i]=c[i]-'0';    }    void add(int p)    {        a[p]++;        ww(a[p]==2)        {            a[p]=0;            a[p-2]++;a[p+1]++;            p-=2;        }        r4(i,400,2)if(a[i]&&a[i-1])a[i]=a[i-1]=0,a[i+1]=1,i=i+3;        l=400,r=-1;        r2(i,2,400)if(a[i]){l=min(l,i),r=max(i,r);}//printf("%d      ",i);}    }};STC two,th;STC add(STC x,STC y){    r2(i,y.l,y.r)if(y.a[i])x.add(i);    return x;}STC s[40];bool ok[31];int main(){    int n;    s[1].get("10.01");    r2(i,2,32)s[i]=add(s[i-1],s[i-1]);    ww(~sf1(n))    {        if(n<=1)pd(n,10);        else        {            clr(ok,0);            r2(i,0,30)if((1<<i)&n)ok[i]=1;            int st;            r2(i,1,30)if(ok[i]){st=i;break;}            STC tm=s[st];            r2(i,st+1,30)if(ok[i]){tm=add(tm,s[i]);}            if(n&1)tm.add(200);            tm.out();puts("");        }    }}

C题
二分

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#define ll long longusing namespace std;ll n,s,t;double p;ll a[2000010],b[2000010];ll d[50];ll get(ll val){    ll ret=0;    for(ll mask=0;mask<(1<<s);mask++)    {        ret+=upper_bound(b,b+(1<<t),val-a[mask])-b;    }    return ret;}int main(){    int cas;scanf("%d",&cas);    while(cas--)    {        scanf("%lld%lf",&n,&p);        for(ll i=1;i<=n;i++)scanf("%lld",&d[i]);        if(p<=0.0)        {            puts("0");continue;        }        s=n/2,t=n-s;        for(ll mask=0;mask<(1<<s);mask++)        {            ll x=0;            for(ll j=0;j<s;j++)if(mask&(1<<j))x+=d[j+1];            a[mask]=x;        }        sort(a,a+(1<<s));a[1<<s]=1e12;        for(ll mask=0;mask<(1<<t);mask++)        {            ll x=0;            for(ll j=0;j<t;j++)if(mask&(1<<j))x+=d[s+j+1];            b[mask]=x;        }        sort(b,b+(1<<t));b[1<<t]=1e12;        ll l=1,r=40000,mid;        while(l<r)        {            mid=l+r>>1;            if(get(mid)/((double)(1LL<<n))<p)l=mid+1;            else r=mid;        }        ll ans=1e16;        for(ll mask=0;mask<(1<<s);mask++)        {            ll pos=lower_bound(b,b+(1<<t),l-a[mask])-b;            ans=min(ans,b[pos]+a[mask]);        }        printf("%lld\n",ans);    }}

F题
题目中说n<=100，其实比100大。
高斯消元

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<vector>using namespace std;int n,m,in[300],g[300][300];vector<int>V[300];namespace G{    const int _=300;    double eps=1e-9;    typedef double ml;    ml a[_][_],X[_];    bool Gauss(ml a[][_],int va,ml X[],int equ,int add=0)    //va是未知量的个数，equ是式子个数,add是增广矩阵    {        for(int r=1;r<=equ;r++)        {            int mx=r;            for(int i=r+1;i<=equ;i++)if(fabs(a[mx][r])<fabs(a[i][r]))mx=i;            if(fabs(a[mx][r])<eps)return 0;            if(mx!=r)for(int i=r;i<=va+1+add;i++)swap(a[mx][i], a[r][i]);            for(int i=r+1;i<=va+1+add;i++)a[r][i]/=a[r][r]; a[r][r]=1;//化简            for(int i=1;i<=equ;i++)if(i!=r)            {                for(int j=r+1;j<=va+1+add;j++)a[i][j]-=a[r][j]*a[i][r];                a[i][r]=0;            }        }        return 1;    }    void out(int n,int m)    {        for(int i=1;i<=n;i++)        {            for(int j=1;j<=m;j++)printf("%lf ",a[i][j]);puts("");        }    }    int sta[300];    void Go()    {        memset(a,0,sizeof a);        memset(X,0,sizeof X);        for(int i=1;i<n;i++)        {            int len=V[i].size();            for(int j=0;j<len;j++)            {                int v=V[i][j];                a[i][v]=1.0/in[v];            }            a[i][i]=-1;a[i][n+1]=0;        }        for(int i=1;i<=n+1;i++)a[n][i]=1;        int add=0;        for(int k=1;k<n;k++)if(g[n][k]==0)        {            sta[++add]=k;            for(int i=1;i<n;i++)            {                a[i][n+1+add]=g[n][i]?1.0/(in[n]+1):0;            }            a[k][n+1+add]=1.0/(in[n]+1);            a[n][n+1+add]=1.0;        }        if(!Gauss(a,n,X,n,add))        {            puts("INF");return;        }        double now=fabs(a[n][n]);        int ans=-1;        for(int i=n+2;i<=n+1+add;i++)        {            if(now>fabs(a[n][i]))                now=fabs(a[n][i]),ans=sta[i-n-1]-1;        }        printf("%d %d\n",1,ans);    }}int main(){    int cas;scanf("%d",&cas);    while(cas--)    {        scanf("%d%d",&n,&m);        memset(g,0,sizeof g);        memset(in,0,sizeof in);        for(int i=1;i<=n;i++)V[i].clear();        while(m--)        {            int x,y;scanf("%d%d",&x,&y);            if(x!=y)g[x+1][y+1]=1;        }        for(int i=1;i<=n;i++)            for(int j=1;j<=n;j++)            {                if(i!=j&&g[i][j])                {                    in[i]++;                    V[j].push_back(i);                }            }        G::Go();    }}

I题
字符串哈希

#include<cstdio>#include<cstring>#include<algorithm>#include<map>#define ll long longusing namespace std;ll mod=1e12+7,t=1331;ll len,m,l,ans;char c[100010];ll sta[100010],top;map<ll,ll>M;ll jie[100010];int main(){    jie[0]=1;    for(ll i=1;i<=100000;i++)jie[i]=jie[i-1]*t%mod;    while(~scanf("%lld%lld",&m,&l))    {        scanf("%s",c+1);        ans=0;        len=strlen(c+1);        for(ll i=1;i<=l&&i+m*l-1<=len;i++)        {            if(i==1)            {                top=0;                ll tmp=0,num=0;                for(ll j=i;j<=len;j++)                {                    ++num;                    tmp=(tmp+c[j]*jie[l-num])%mod;                    if(num==l)                    {                        sta[++top]=tmp;                        tmp=0;num=0;                    }                }            }            else            {                top=(len-i+1)/l;                for(ll j=1;j<=top;j++)                {                    sta[j]-=c[i-1+l*(j-1)]*jie[l-1]%mod;                    sta[j]=(sta[j]%mod+mod)%mod;                    sta[j]=sta[j]*t%mod;                    sta[j]=(sta[j]+c[i+l*j-1])%mod;                }            }            ll ban=0;M.clear();            for(ll j=1;j<=top;j++)            {                if(M[sta[j]]&&j-M[sta[j]]<m)                {                    ban=max(ban,M[sta[j]]+m);                }                M[sta[j]]=j;                if(j>=m&&ban<=j)ans++;            }        }        printf("%lld\n",ans);    }}

J题
LCA问题，树上三个点，问各个点统治的点个数。

#include<cstdio>#include<cstring>#include<algorithm>#include<vector>using namespace std;int n,m;namespace G{    const int _=100010;    vector<int>V[_];    int fa[_][21],dep[_],siz[_];    void dfs(int u,int f)    {        for(int i=1;i<=20;i++)fa[u][i]=fa[fa[u][i-1]][i-1];        int len=V[u].size();        siz[u]=1;        for(int i=0;i<len;i++)        {            int v=V[u][i];            if(v==f)continue;            fa[v][0]=u;            dep[v]=dep[u]+1;            dfs(v,u);            siz[u]+=siz[v];        }    }    void init()    {        fa[1][0]=0;dep[1]=1;        dfs(1,0);    }    int fa_k(int u,int k)    {        for(int i=0;i<=20;i++)if(k&(1<<i))u=fa[u][i];return u;    }    int query(int a,int b)    {        if(dep[a]>dep[b])swap(a,b);        b=fa_k(b,dep[b]-dep[a]);        if(a==b)return a;        for(int i=20;i>=0;i--)        {            if(fa[a][i]!=fa[b][i])            {                a=fa[a][i];b=fa[b][i];            }        }        return fa[a][0];    }    struct STC{        int flag,u;    };    STC getMid(int a,int b,int ab)    {        int tmp=dep[a]+dep[b]-2*dep[ab];        STC t;        if(dep[a]>=dep[b])            t.flag=1,t.u=fa_k(a,(tmp-1)/2);        else            t.flag=2,t.u=fa_k(b,tmp/2);        return t;    }    int cal(int a,int b,int c,int ab,int ac)    {        STC abm=getMid(a,b,ab),acm=getMid(a,c,ac);        if(abm.flag==1&&acm.flag==1)            return dep[abm.u]>dep[acm.u]?siz[abm.u]:siz[acm.u];        else if(abm.flag==2&&acm.flag==2)        {            if (dep[abm.u]<dep[acm.u]) swap(abm,acm);            if (query(abm.u,acm.u)==acm.u) return n-siz[acm.u];            return n-siz[abm.u]-siz[acm.u];        }        else        {            if(abm.flag==2)swap(abm,acm);            int tmp=query(abm.u,acm.u);            if(tmp==abm.u)return siz[abm.u]-siz[acm.u];            else return siz[abm.u];        }    }    void Go()    {        int cas;scanf("%d",&cas);        while(cas--)        {            scanf("%d",&n);            for(int i=1;i<=n;i++)V[i].clear();            for(int i=1;i<n;i++)            {                int x,y;scanf("%d%d",&x,&y);                V[x].push_back(y);V[y].push_back(x);            }            init();            scanf("%d",&m);            while(m--)            {                int a,b,c;                scanf("%d%d%d",&a,&b,&c);                int ab=query(a,b),ac=query(a,c),bc=query(b,c);                printf("%d %d %d\n",cal(a,b,c,ab,ac),cal(b,a,c,ab,bc),cal(c,a,b,ac,bc));            }        }    }};int main(){    G::Go();}