zoj 3547 The Boss on Mars

需要用到概率论的容斥定理以及计算1 ^ 4 + 2 ^ 4 + ……+ n ^ 4的计算公式1^4+2^4+……+n^4=n(n+1)(2n+1)(3n^2+3n-1)/30

 

#pragma comment(linker,"/STACK:1024000000,1024000000")

#include <cstdio>

#include <cmath>



#define LL long long

const LL mod = 1e9 + 7;

#define MAX 10000



int len, prime[MAX], num[30];

bool vis[MAX + 5];

LL n, sum, pi;



void get_prime(){

	len = 0;

	for(int i = 2; i<=MAX; ++i){

		if(!vis[i]) prime[len++] = i;

		for(int j = i * i; j <= MAX; j+=i) vis[j] = 1;

	}

}



LL power(LL x, LL y){

	if(y == 0) return 1;

	if(y == 1) return x;

	LL v = power(x, y / 2);

	v = v * v % mod;

	if(y % 2 == 1) v = v * x % mod;

	return v;

}



LL cal(LL v){

	return v * (v  + 1) % mod * (v * 2 + 1) % mod * (v * v * 3 % mod + v * 3 - 1 + mod) % mod * pi % mod;

}



void dfs(int cnt, int p, int pos, LL s){

	if(cnt % 2 == 1) sum = (sum + cal(n / s) * s % mod * s % mod * s % mod * s % mod) % mod;

	else sum = (sum - cal(n / s) * s % mod * s % mod * s % mod * s % mod + mod) % mod;

	for(int i = pos; i < p; ++i)

		dfs(cnt + 1, p, i + 1, s * num[i] % mod);

}



int main ()

{

	//freopen ("in.txt", "r", stdin);

	get_prime();

	//for(int i = 0; i < len; ++i) printf("%d ", prime[i]);

	pi = power(30, mod - 2);

	int t;

	scanf ("%d", &t);

	while (t--)

	{

		int x, p = 0;

		scanf("%d", &x);

		n = x;

		sum = cal(n);

		//printf("%d\n", n);

		for(int i = 0; i < len; ++i){

			int v = prime[i];

			if(v > x) break;

			if(x % v == 0) num[p++] = v;

			while(x % v ==0) x /= v;

		}

		if(x > 1) num[p++] = x;

		//for(int i = 0; i < p; ++i) printf("%d ", num[i]);

		for (int i = 0; i < p; ++i)

			dfs(0, p, i + 1, (LL)num[i]);

		printf("%lld\n", sum);

	}

	return 0;

}