2017 ACM/ICPC Asia Regional Xi'an Online

A Tree

unsolved

B Coin

分别加减得到的二次项系数做差即可剩余偶数项

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const LL mod=1e9+7;

LL pow_mod(LL a,LL n){
    long long res=1,t=a;
    while(n){
        if(n&1) res=(res*t)%mod;
        n/=2;
        t=(t*t)%mod;
    }
    return res;
}

int main(){
    LL t;
    scanf("%lld",&t);
    while(t--){
        LL p,q,k;
        scanf("%lld%lld%lld",&p,&q,&k);
        LL x=(2*q-p);
        x=(x%mod+mod)%mod;
        x=(LL)(pow_mod(x,k)*pow_mod(pow_mod(p,k),mod-2))%mod;
        if(k%2==0) x++;
        else x=1-x;
        x=(x%mod+mod)%mod;
        x=(LL)x*pow_mod(2,mod-2)%mod;
        printf("%lld
",x);
    }


    return 0;
}

C Sum

输出233个10000000

#include <stdio.h>
#include <iostream>
using namespace std;
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        int x;
        scanf("%d",&x);
        for(int i=1;i<=233;i++){
            printf("10000000");
        }
        printf("
");
    }

    return 0;
}

D Brain-baffling Game

unsolved

E Maximum Flow

每个数的贡献为min（x，x ^（n-1）），暴力找0和1出现的规律。

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const LL mod=1e9+7;

LL ans1,ans2;

LL fun(LL x){
    LL cnt=0;
    while(x){
        x/=2;
        cnt++;
    }
    return cnt-1;
}

LL p[65];

int main(){
    p[0]=1;
    for(LL i=1;i<65;i++) p[i]=p[i-1]*2%mod;
    LL n;
    while(scanf("%lld",&n)!=EOF){
        n--;
        LL tttt=n;
        ans1=ans2=0;
        LL cnt=fun(n);
        LL tmp=(1LL<<cnt)-1;
        tmp%=mod;
        ans1=(tmp*(tmp+1)/2)%mod;

        n=n^(1LL<<cnt);
        LL res=0;
        for(LL i=0;i<=cnt;i++){
            LL num=0;
            res+=p[i]*((n&(1LL<<i))?1:0);
            res%=mod;
            num+=(n>>(i+1))*p[i]%mod;
            if(n&(1LL<<i)) num+=res-p[i]+1;
            if(n&(1LL<<i)) ans2+=((n-num+1)%mod+mod)*p[i]%mod;
            else ans2+=(num%mod)*p[i]%mod;
            ans2%=mod;
        }
        LL ans=(ans1+ans2+tttt)%mod;
        printf("%lld
",ans);
    }

    return 0;
}

F Trig Function

好像有公式。

G Xor

Big-Small

1.对于小数据，预处理出每个点向上跳，跳到根，步伐为k的结果。

2.对于大数据，两个端点都倍增得向上跳到lca。

#include <bits/stdc++.h>
using namespace std;

const int MAX_V=5e4+5;
const int MAX_LOG=18;
int S;

int n,q;
int a[MAX_V];

vector<int> g[MAX_V];
int fa[MAX_LOG][MAX_V];
int depth[MAX_V];
void addedge(int u,int v){
    g[u].push_back(v);
    g[v].push_back(u);
}
void dfs(int u,int pre,int d){
    fa[0][u]=pre;
    depth[u]=d;
    for(int i=0;i<g[u].size();i++){
        int v=g[u][i];
        if(v==pre) continue;
        dfs(v,u,d+1);
    }
}
void init(int n){
    depth[0]=0;
    dfs(1,0,1);
    for(int k=0;k+1<MAX_LOG;k++){
        fa[k+1][0]=fa[k][0]=0;
        for(int v=1;v<=n;v++){
            fa[k+1][v]=fa[k][fa[k][v]];
        }
    }
}
int LCA(int u,int v){
    if(depth[u]>depth[v]) swap(u,v);
    for(int k=MAX_LOG-1;k>=0;k--){
        if((depth[v]-depth[u])&(1<<k)){
            v=fa[k][v];
        }
    }
    if(u==v) return u;
    for(int k=MAX_LOG-1;k>=0;k--){
        if(fa[k][u]!=fa[k][v]){
            u=fa[k][u];
            v=fa[k][v];
        }
    }
    return fa[0][v];
}

int id[MAX_V];
int dp[MAX_V][605];
int par[MAX_V][605];
void dfs2(int u,int d){
    id[d]=u;
    par[u][0]=u;
    for(int i=1;i<=S;i++){
        dp[0][i]=0;
        if(d<=i){
            dp[u][i]=a[u];
            par[u][i]=0;
        }
        else {
            dp[u][i]=a[u]^dp[id[d-i]][i];
            par[u][i]=id[d-i];
        }
    }
    for(int i=0;i<g[u].size();i++){
        int v=g[u][i];
        if(v==fa[0][u]) continue;
        dfs2(v,d+1);
    }
}

int pre(int u,int k){
    int res=u;
    for(int i=MAX_LOG-1;i>=0;i--){
        if(k&(1<<i)) res=fa[i][res];
    }
    return res;
}

int solve1(int u,int v,int k){
    int lca=LCA(u,v);
    int res=(depth[v]+depth[u]-2*depth[lca])%k;
    int s=u,t=pre(v,res);
    int ans=0;
    if((depth[u]-depth[lca])%k==0) ans=a[lca];
    while(depth[s]>depth[lca]){
        ans^=a[s];
        s=pre(s,k);
    }
    while(depth[t]>depth[lca]){
        ans^=a[t];
        t=pre(t,k);
    }
    return ans;
}
int solve2(int u,int v,int k){
    int lca=LCA(u,v);
    int res1=(k-((depth[u]-depth[lca])%k))%k;
    int s1=u,s2=par[lca][res1];

    int ans=dp[s1][k]^dp[s2][k];
    if((depth[u]-depth[lca])%k==0) ans^=a[lca];

    int res2=(depth[v]+depth[u]-2*depth[lca])%k;
    int t1=par[v][res2];
    if(lca!=v&&depth[t1]>depth[lca]){
        int res3=(k-((depth[t1]-depth[lca])%k))%k;
        int t2=par[lca][res3];
        ans^=(dp[t1][k]^dp[t2][k]);
    }

    return ans;
}

int main(){
    while(~scanf("%d%d",&n,&q)){
        S=min(600,(int)sqrt(q*(log(1.0*n)/log(2.0))));
        for(int i=1;i<=n;i++){
            g[i].clear();
        }
        for(int i=1;i<n;i++){
            int u,v;
            scanf("%d%d",&u,&v);
            addedge(u,v);
        }
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
        }
        init(n); dfs2(1,1);
        while(q--){
            int u,v,k;
            scanf("%d%d%d",&u,&v,&k);
            if(k>S) printf("%d
",solve1(u,v,k));
            else printf("%d
",solve2(u,v,k));
        }
    }

    return 0;
}

H Music

unsolved

I Barty's Computer

unsolved

J Easy Problem

unsolved