uva 10692(指数循环节)

递归用的很巧妙，虽然不是我想出来的

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#define ls 2*rt
#define rs 2*rt+1
#define lson ls,L,mid
#define rson rs,mid+1,R
#define ll long long
using namespace std;
typedef pair<int,int> pii;
const ll inf = 0x3f3f3f3f;
/*void dis(int a[], int n){
	printf("总数为%d个\n",n); 
	for(int i = 0; i < n; i++) 	cout<<a[i]<<", ";
	cout<<endl<<"------------------"<<endl;		
}*/
const int mx = 1e3+100;
char s[mx];
int a[mx],mod,n;

ll eular(ll n)
{
    ll ret=1,i;
    for(i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            n/=i,ret*=i-1;
            while(n%i==0) n/=i,ret*=i;
        }
    }
    if(n>1) ret*=n-1;
    return ret;
}
ll power(ll a, ll b, ll m){
	if(b == 0)
	 return 1;
	 ll  ans= 1;
	 a %= m;
	 while(b){
	   if(b%2)
	       ans = ans*a%m;
	    b/=2;
	    a = a*a%m;
	 }
	 
	 return ans;
} 
ll slove(int i, int mod){
    if(i == n-1)
    	return a[i]%mod;
    int eu = eular(mod);
    int nu = slove(i+1,eu) + eu;
	return power(a[i],nu,mod);
}

int main(){
	int ca = 0;
	
	while(scanf("%s",s) && s[0] != '#'){
		mod = 0;
		int len = strlen(s);
		for(int  i = 0; i < len; i++)
			mod = mod*10+s[i]-'0';
		scanf("%d",&n);
		for(int i = 0; i < n; i++){
			scanf("%d",a+i);			
		}
		printf("Case #%d: %lld\n",++ca,slove(0,mod));

	}

	
	return 0;
}