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;
}