[UVA10692]Huge Mods

题目链接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1633

求一连串的幂，如果暴力算的话会溢出，题中会有一个求模，给出一个欧拉定理的应用：

x^y % m = x^(y % phi[m] + phi[m]) % m

递归求解即可。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

const int maxn = 1111111;
int mod, n, x;
int a[maxn];
char str[maxn];
int phi[maxn];

int quickmul(int x, int n, int mod) {
    int ans = 1;
    int t = x;
    while(n) {
        if(n & 1) {
            ans = (ans * t) % mod;
        }
        t = t * t % mod;
        n >>= 1;
    }
    return ans;
}

void geteular() {
    memset(phi, 0, sizeof(phi));
    phi[1] = 1;
    for(int i = 2; i < maxn; i++) {
        if(!phi[i]) {
            for(int j = i; j < maxn; j+=i) {
                if(!phi[j]) {
                    phi[j] = j;
                }
                phi[j] = phi[j] / i * (i - 1);
            }
        }
    }
}

int solve(int x, int m) {
    if(x == n - 1) {
        return a[x] % m;
    }
    return quickmul(a[x], solve(x+1, phi[m]) + phi[m], m);
}

int main() {
    geteular();
    int kase = 1;
    while(~scanf("%s", str) && strcmp(str, "#")) {
        sscanf(str, "%d", &mod);
        scanf("%d", &n);
        for(int i = 0; i < n; i++) {
            scanf("%d", &a[i]);
        }
        int ans = solve(0, mod);
        printf("Case #%d: %d
", kase++, ans);
    }
}