hdu 4059 The Boss on Mars-容斥定理-四次方求和_综合

题目链接：This is the link

The Boss on Mars

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3290 Accepted Submission(s): 1082

Problem Description

On Mars, there is a huge company called ACM (A huge Company on Mars), and it’s owned by a younger boss.

Due to no moons around Mars, the employees can only get the salaries per-year. There are n employees in ACM, and it’s time for them to get salaries from their boss. All employees are numbered from 1 to n. With the unknown reasons, if the employee’s work number is k, he can get k^4 Mars dollars this year. So the employees working for the ACM are very rich.

Because the number of employees is so large that the boss of ACM must distribute too much money, he wants to fire the people whose work number is co-prime with n next year. Now the boss wants to know how much he will save after the dismissal.

Input

The first line contains an integer T indicating the number of test cases. (1 ≤ T ≤ 1000) Each test case, there is only one integer n, indicating the number of employees in ACM. (1 ≤ n ≤ 10^8)

Output

For each test case, output an integer indicating the money the boss can save. Because the answer is so large, please module the answer with 1,000,000,007.

Sample Input

2 4 5

Sample Output

82 354

Hint

Case1: sum=1+3*3*3*3=82 Case2: sum=1+2*2*2*2+3*3*3*3+4*4*4*4=354

Author

ZHANG, Chao

Source

2011 Asia Dalian Regional Contest

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题目大意：

给定 T 组数据，每组数据一个数 n ，然后让你求 <n且与 n 互素的数的四次方的和，结果对1000000007取模输出

思路：

就是求一个 S = a1^4 + a2^4 + ... + ak^4;，ai表示与n互素的数

直接枚举计算，要超时，所以要使用容斥定理计算，

需要推导四次方求和公式： $sum=\frac{n*(n+1)*(2*n+1)(3*n*n+3*n-1)}{30}$ //尝试自己推一下

同时这里需要用逆元计算求模，所以需要计算30关于1000000007的逆元（233333335）

注意：还要自己写快速幂求模

//二进制求容斥定理

#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include <iomanip>
#include<list>
#include<map>
#include<queue>
#include<sstream>
#include<stack>
#include<string>
#include<set>
#include<vector>
using namespace std;
#define PI acos(-1.0)
#define EPS 1e-8
//#define MOD 1e9+7
#define LL long long
#define ULL unsigned long long     //1844674407370955161
#define INT_INF 0x7f7f7f7f      //2139062143
#define LL_INF 0x7f7f7f7f7f7f7f7f //9187201950435737471
const int dr[]={0, 0, -1, 1, -1, -1, 1, 1};
const int dc[]={-1, 1, 0, 0, -1, 1, -1, 1};
// ios::sync_with_stdio(false);
// 那么cin, 就不能跟C的 scanf，sscanf, getchar, fgets之类的一起使用了。
vector<LL >  fac;
const LL mod=1e9+7;
LL inv;//30关于mod的逆元
void Fac(LL n)//分解素因子
{fac.clear();for(LL i=2;i*i<=n;++i){if(!(n%i)){fac.push_back(i);while(!(n%i))n/=i;}}if(n>1)fac.push_back(n);
}
LL Quick_pow(LL a,LL n)//快速幂求模
{LL ret=1;while(n){if(n&1)ret=ret*a%mod;n>>=1;a=a*a%mod;}return ret%mod;
}
LL Sum(LL n)//四次方求和
{LL ret=n%mod;ret=ret*(n+1)%mod;ret=ret*(n+n+1)%mod;ret=ret*((3*n*n+3*n-1)%mod)%mod;ret=ret*inv%mod;return ret;
}
void Solve(LL n)//容斥定理
{Fac(n);LL sum=fac.size();LL ret=0;for(int i=1;i<(1<<sum);++i){LL ans=1;LL cnt=0;for(int j=0;j<sum;++j){if(i&(1<<j)){ans=ans*fac[j]%mod;++cnt;}}if(cnt&1)ret=(ret+Sum(n/ans)*Quick_pow(ans,4)+mod)%mod;elseret=(ret-Sum(n/ans)*Quick_pow(ans,4)+mod)%mod;}ret=(Sum(n)-ret+mod)%mod;ret=(ret%mod+mod)%mod;printf("%lld\n",ret);
}
int main()
{inv=Quick_pow(30,mod-2);//计算30关于mod的逆元int t;scanf("%d",&t);while(t--){LL n;scanf("%lld",&n);Solve(n);}return 0;
}