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Problem X

Huge Mod

Input: standard input

Output: standard output

Time Limit: 1 second

The operator for exponentiation is different from the addition, subtraction, multiplication or division operators in the sense that the default associativity for exponentiation goes right to left instead of left to right. So unless we mess it up by placing parenthesis, should mean not . This leads to the obvious fact that if we take the levels of exponents higher (i.e., 2^3^4^5^3), the numbers can become quite big. But let‘s not make life miserable. We being the good guys would force the ultimate value to be no more than 10000.

Input

Output

10 4 2 3 4 5
100 2 5 2
53 3 2 3 2
#

Case #1: 2
Case #2: 25
Case #3: 35

Problem setter: Monirul Hasan, Member of Elite Problemsetters‘ Panel

Special thanks: Mohammad Sajjad Hossain

题意：输入正整数a1,a2,a3..an和模m，求a1^a2^...^an mod m

思路：需要用到一个公式：，递归处理

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
typedef long long ll;
using namespace std;
const int maxn = 1010;

int a[maxn], vis[maxn], n;
char str[maxn];

int euler_phi(int m) {
	int tmp = (int) sqrt(m+0.5);
	int ans = m;
	for (int i = 2; i <= tmp; i++) if (m % i == 0) {
		ans = ans / i * (i-1);
		while (m % i == 0)
			m /= i;
	}
	if (m > 1)
		ans = ans / m * (m-1);
	return ans;
}

int pow_mod(int a, int m, int mod) {
	int tmp = 1;
	while (m)  {
		if (m & 1)
			tmp = tmp * a % mod;
		m >>= 1;
		a = a * a % mod;
	}
	return tmp;
}

int solve(int cur, int m) {
	if (cur == n-1)
		return a[cur] % m;
	int phi_m = euler_phi(m);
	int tmp = solve(cur+1, phi_m);
	return pow_mod(a[cur], tmp + phi_m, m);
}

int main() {
	int cas = 1, mod;
	while (scanf("%s", str) != EOF && str[0] != '#') {
		mod = 0;
		for (int i = 0; str[i]; i++)
			mod = mod*10 + str[i] - '0';
		scanf("%d", &n);
		for (int i = 0; i < n; i++)
			scanf("%d", &a[i]);
		printf("Case #%d: %d\n", cas++, solve(0, mod));
	}
	return 0;
}

UVA - 10692 Huge Mods （欧拉函数）

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原文地址：http://blog.csdn.net/u011345136/article/details/38817775