2016 Multi-University Training Contest 1 1006 PowMod

題目鏈接：點擊打開鏈接

題目大意：！！！！！！！很嚇人的數學取模

解題思路：這次真的不好意思掛原創啊，因爲完全是按照官方題解來打的，數學差沒辦法。

以上是官方題解給出的公式：官方題解

代碼：

#include<iostream>
#include<vector>
#include<cmath>
#include<algorithm>
#include<ctime>
#include "cstdio"
#include "string"
#include "string.h"
#include "map"
using namespace std;
const int maxn = 10000001;
const int mod = 1000000007;
int vis[maxn], phi[maxn], prime[maxn], pt, a[20];
long long sum[maxn];
void init()
{
	pt = 0;
	phi[1] = 1;
	int N = maxn;
	int k;
	for (int i = 2;i<N;i++)
	{
		if (!vis[i])
			prime[pt++] = vis[i] = i, phi[i] = i - 1;
		for (int j = 0;j<pt && (k = prime[j] * i)<N;j++)
		{
			vis[k] = prime[j];
			if (vis[i] == prime[j])
			{
				phi[k] = phi[i] * prime[j];
				break;
			}
			else
				phi[k] = phi[i] * (prime[j] - 1);
		}
	}
}
long long pow(long long a, long long n, long long mod)
{
	long long base = a, ret = 1;
	while (n)
	{
		if (n % 2)
			ret = (ret*base) % mod;
		base = (base*base) % mod;
		n /= 2;
	}
	return ret;
}
long long cal_sum(int pos, long long n, long long m)
{
	if (n == 1)//sum(1,m)=sigma(phi(i))
		return sum[m];
	if (m == 0)//sum(n,0)=0;
		return 0;
	return ((a[pos] - 1)*cal_sum(pos - 1, n / a[pos], m) % mod + cal_sum(pos, n, m / a[pos])) % mod;
	//sum(n,m)=phi(p)*sum(n/p,m)+sum(n,m/p),p爲質數
}
long long cal_ans(long long k, long long mod)
{
	if (mod == 1)
		return 0;
	long long temp = cal_ans(k, phi[mod]);//b%phi(p)
	long long ans = (pow(k,temp+phi[mod],mod)+mod)%mod;
	return ans;
}
int main()
{
	init();
	sum[0] = 0;
	for (int i = 1;i < maxn;i++)
		sum[i] = (sum[i - 1] + phi[i]) % mod;
	int n, m, p;
	while (scanf("%d %d %d", &n, &m, &p) != EOF)
	{
		int now = n, num = 0;
		for (int i = 0;i < pt;i++)
		{
			if (!vis[n])//n是質數
			{
				a[num++] = now;
				break;
			}
			if (now%prime[i] == 0)//分解質因數
			{
				a[num++] = prime[i];
				now /= prime[i];
			}
		}
		long long temp = cal_sum(num - 1, n, m);
		long long ans = cal_ans(temp, p);
		printf("%lld\n", ans);
	}
	return 0;
}