HDU 5525：Product 欧拉定理

Product

 

 Accepts: 21

 

 Submissions: 171

 Time Limit: 6000/3000 MS (Java/Others)

 

 Memory Limit: 131072/131072 K (Java/Others)

问题描述

给n个数{A}_{1},{A}_{2}....{A}_{n}A​1​​,A​2​​....A​n​​，表示N=\prod_{i=1}^{n}{i}^{{A}_{i}}N=∏​i=1​n​​i​A​i​​​​。求N所有约数之积。

输入描述

输入有多组数据.
每组数据第一行包含一个整数n.(1\leq n\leq {10}^{5})(1≤n≤10​5​​)
第二行n个整数{A}_{1},{A}_{2}....{A}_{n}A​1​​,A​2​​....A​n​​,保证不全为0.(0\leq {A}_{i}\leq {10}^{5})(0≤A​i​​≤10​5​​).
数据保证 $\sum n\le 500000$.

输出描述

对于每组数据输出一行为答案对{10}^{9}+710​9​​+7取模的值.

输入样例

4
0 1 1 0
5
1 2 3 4 5

输出样例

36
473272463

官方题解：

做了一道div1的第三题，很好的一道题

参考了别人的代码，其中一处是(p^index)%mod，如果index很大的话，根据欧拉定理，就可以变成(p^(index%(phi(mod))))%mod。记录一下。

代码：

#pragma warning(disable:4996)  
#include <iostream>  
#include <algorithm>  
#include <cmath>  
#include <vector>  
#include <string>  
#include <cstring>  
using namespace std;

typedef long long ll;

ll N;
const int maxn = 100010;
const ll mod = 1e9 + 7;
const ll mod2 = 2 * (mod - 1);
ll index[maxn], L[maxn], R[maxn];
int num, pri[maxn], vis[maxn];
vector<int>have[maxn];

void init()
{
	num = 0;
	int i, j, n;
	for (i = 2; i < maxn; i++)
	{
		if (vis[i])
			continue;
		pri[++num] = i;
		for (j = i + i; j < maxn; j = j + i)
		{
			vis[j] = 1;
		}
	}
	for (i = 1; i < maxn; i++)
	{
		n = i;
		for (j = 1; j <= num&&pri[j] <= n; j++)
		{
			while (n%pri[j] == 0)
			{
				have[i].push_back(j);//记录所含有的质数，用质数的下标记录
				n /= pri[j];
			}
		}
	}
}

ll getresult(ll A, ll n, ll k)
{
	ll b = 1;
	while (n > 0)
	{
		if (n & 1)
		{
			b = (b*A) % k;
		}
		n = n >> 1;
		A = (A*A) % k;
	}
	return b;
}

void solve(int cishu, int n)
{
	int temp;
	int si = have[cishu].size();
	for (int i = 0; i < si; i++)
	{
		temp = have[cishu][i];
		index[temp] = (index[temp] + n) % mod2;
	}
}

int main()
{
	//freopen("i.txt", "r", stdin);
	//freopen("o.txt", "w", stdout);

	int x;
	ll k, n, ans;
	init();
	while (scanf("%d", &N) == 1)
	{
		memset(index, 0, sizeof(index));
		for (int i = 1; i <= N; i++)
		{
			scanf("%d", &x);
			solve(i, x);
		}
		int p = 1;
		while (pri[p] < N)
			p++;
		N = p;
		L[0] = R[N + 1] = 1;
		for (int i = 1; i <= N; i++)
			L[i] = L[i - 1] * (index[i] + 1) % mod2;
		for (int i = N; i >= 1; i--)
			R[i] = R[i + 1] * (index[i] + 1) % mod2;
		ans = 1;
		for (int i = 1; i <= N; i++)
		{
			k = L[i - 1] * R[i + 1] % mod2;
			n = index[i] * (index[i] + 1) / 2 % mod2;
			ans = ans*getresult(pri[i], n*k%mod2,mod) % mod;
		}
		printf("%lld\n", ans);
	}
	//system("pause");
	return 0;
}