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.

Given a₁, a₂, a₃, ... , a_N and m(=10000)

you only need to compute a₁^a₂^a₃^...^a_N mod m.

Input

There can be multiple (not more than 100) test cases. Each test case will be presented in a single line. The first line of each test case would contain the value for M(2<=M<=10000). The next number of that line would be N(1<=N<=10). Then N numbers - the values
for a₁, a₂, a₃, ... , a_N would follow. You can safely assume that 1<=a_i<=1000. The end of input is marked by a line containing a single hash (‘#‘) mark.

Output

For each of the test cases, print the test case number followed by the value of a₁^a₂^a₃^...^a_N mod m on one line. The sample output shows the exact format for printing the test case number.

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;
}