2017 Multi-University Training Contest

TrickGCD

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2097    Accepted Submission(s): 816

Sample Input

1

4

4 4 4 4

Sample Output

Case #1: 17

 

Source

2017 Multi-University Training Contest - Team 2

最近不舒服 病了 找个时间来解释

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<string.h>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<cmath>
typedef long long ll;
typedef unsigned long long LL;
using namespace std;
const double PI=acos(-1.0);
const double eps=0.0000000001;
const int N=200000+100;
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
int num[N],mo[N];
void mobius(){
    mo[1]=1;
    for(int i=1;i<=N;i++)
    for(int j=i+i;j<=N;j+=i)
    mo[j]-=mo[i];
}
int pow_mod(int a,int b){
    ll ans=1;
    while(b)
    {
        if(b&1)ans=(ans*a)%mod;
        a=(1ll*a*a)%mod;
        b>>=1;
    }
    return ans;
}
int main(){
    int t,n,x,minn,kase=0;
    mobius();
    scanf("%d",&t);
    while(t--)
    {
        memset(num,0,sizeof(num));
        for(int i=1;i<=N;i++){
            num[i]=num[i-1]+num[i];
        }
        minn=INF;
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%d",&x);
            minn=min(minn,x);
            num[x]++;
        }
        ll ans=0;
        for(int i=2;i<=minn;i++)
        {
            int cnt=1;
            for(int j=1;j*i<=100000;j++)
            {
                cnt=(1(ll)*cnt*pow_mod(j,num[i*(j+1)-1]-num[i*j-1]))%mod;
            }
            ans=(ans-1(ll)*mo[i]*cnt+mod)%mod;
        }
        printf("Case #%d: %lld
",++kase,ans);
    }
    return 0;
}