HDU - 4059 The Boss on Mars 容斥

题目：求1-n中与n互质的数的4次方之和

思路：容斥定理。

要用到公式：(1^4+2^4+……+n^4)=(n*(n+1)*(2n+1)*(3*n*n+3*n-1))/30

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<algorithm>
#include<ctime>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<list>
#include<numeric>
using namespace std;
#define LL long long
#define ULL unsigned long long
#define INF 0x3f3f3f3f3f3f3f3f
#define mm(a,b) memset(a,b,sizeof(a))
#define PP puts("*********************");
template<class T> T f_abs(T a){ return a > 0 ? a : -a; }
template<class T> T gcd(T a, T b){ return b ? gcd(b, a%b) : a; }
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
// 0x3f3f3f3f3f3f3f3f

const LL MOD=1000000007;
const int maxn=1e5+50;
bool isprime[maxn];
int prime[maxn],tol;
LL Inv,sum;
int fac[maxn],cnt;
void make_prime(int n){
    for(int i=0;i<=n;i++)
        isprime[i]=true;
    isprime[0]=isprime[1]=false;
    tol=0;
    for(int i=2;i<=n;i++){
        if(isprime[i])
            prime[tol++]=i;
        for(int j=0;j<tol;j++){
            if(i*prime[j]<=n)
                isprime[i*prime[j]]=false;
            else
                break;
            if(i%prime[j]==0)
                break;
        }
    }
}
LL pow_mod(LL a,LL b){
    LL res=1;
    while(b){
        if(b&1) res=res*a%MOD;
        a=a*a%MOD;
        b/=2;
    }
    return res;
}
LL pow4(LL n){
    return n*n%MOD*n%MOD*n%MOD;
}
LL get_sum(LL n){
    LL res=1;
    res=res*n%MOD*(n+1)%MOD;
    res=res*((2*n+1)%MOD)%MOD;
    res=res*((3*n*n%MOD+3*n%MOD-1)%MOD)%MOD;
    res=res*Inv%MOD;
    res=(res+MOD)%MOD;
    return res;
}
void dfs(int pos,int x,int num,LL n){
    if(pos==cnt){
        if(num&1){
            sum-=get_sum(n/x)*pow4((LL)x)%MOD;
            sum=(sum+MOD)%MOD;
        }
        else{
            sum+=get_sum(n/x)*pow4((LL)x)%MOD;
            sum=(sum+MOD)%MOD;
        }
        return;
    }
    dfs(pos+1,x*fac[pos],num+1,n);
    dfs(pos+1,x,num,n);
}
int main(){

    int T;
    LL n;
    make_prime(maxn-5);
    Inv=pow_mod((LL)30,MOD-2);
    scanf("%d",&T);
    while(T--){
        scanf("%lld",&n);
        cnt=0;
        LL m=n;
        for(int i=0;i<tol&&prime[i]*prime[i]<=m;i++){
            if(m%prime[i]==0){
                fac[cnt++]=prime[i];
                while(m%prime[i]==0)
                    m/=prime[i];
            }
        }
        if(m>1)
            fac[cnt++]=m;
        sum=0;
        dfs(0,1,0,n);
        printf("%lld\n",sum);
    }
    return 0;
}