题意:
输入正整数a1,a2,a3..an和模m,求a1^a2^...^an mod m
解析:
#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _ ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = , INF = 0x7fffffff;
LL A[maxn], num[maxn];
LL n;
char str[maxn];
LL qpow(LL a, LL b, LL m)
{
LL res = ;
while(b)
{
if(b & ) res = res * a % m;
a = a * a % m;
b >>= ;
}
return res;
} void init()
{
for(int i=; i<maxn; i++)
A[i] = i;
for(int i=; i<maxn; i++)
if(A[i] == i)
for(int j=i; j<maxn; j+=i)
A[j] = A[j]/i*(i-);
} LL dfs(LL cnt, LL m)
{
if(cnt == n-)
{
return num[cnt] % m;
}
LL phi = A[m];
LL k = dfs(cnt+, phi) + phi; //因为在上一步的快速幂中已经%phi 所有这一步不用%phi
return qpow(num[cnt], k, m);
} int main()
{
init();
int kase = ;
while(scanf("%s",str) && strcmp(str, "#"))
{
LL MOD;
sscanf(str,"%lld", &MOD);
cin>> n;
for(int i=; i<n; i++)
{
cin>> num[i];
}
printf("Case #%d: %lld\n",++kase,dfs(, MOD));
}
return ;
}