（A^B)%C 快速幂乘

<span style="font-size:18px;"><strong>简单数论的快速幂乘问题</strong></span>
#include<iostream>
using namespace std;
int main()
{
int a,b,c;
while (scanf("%d%d%d",&a,&b,&c)!=EOF&&a!=0&&b!=0&&c!=0)
{
int ans = 1;
a=a%c;
while (b>0)
{
if(b%2==1)     // 作用1：当 b为奇数，则先单独乘一个a
// 作用：当 b=1时，即 已经乘了 b=b/2=1后，将值赋给k
ans = (ans * a) % c;
a = (a * a) % c;
b = b / 2;       //a1=a%c                             1=2^0
// a2=((a%c)*(a%c))%c                2=2^1
// a3=((a%c*a%c)%c*(a%c*a%c)%c)      4=2^2
//由此，可知 b=b/2 ,每次的 a 的个数为上一次的 2倍
}
printf("%d\n",ans);
}
return 0;
}

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 2005001;
int g[maxn];
unsigned long long a,b;
int PowerMod(unsigned long long a,unsigned long long b,int c)
{
int ans = 1;
a = a % c;
while(b>0)
{
if(b % 2 == 1)
ans = (ans * a) % c;
a = (a * a) % c;
b = b / 2;
}
return ans;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n;
cin >> a >> b >> n;
if(n == 1)
{
puts("0");
continue;
}
int m = 1;
g[1] = 1,g[2] = 1;
int i = 3;
while(1)
{
g[i] = (g[i - 2] + g[i - 1]) % n;
if(g[1] == g[i-1] && g[2] == g[i])
break;
i++;
}
m = i - 2;
int s = PowerMod(a,b,m);
cout<<g[s]<<endl;
}
return 0;
}