a^b.cpp

传送门

code

#include
using namespace std;
int a,d,c;

int main()
{
	cin>>a>>d>>c;
	
	int ans=1,now=a,b=d;
	while(b>0)//快速幂 
	{
		if(b&1)
		{
			ans=(long long)ans*now%c;//long long防溢出 
		}
		now=(long long)now*now%c;//long long防溢出 
		b/=2;
	}
	
	printf("%d^%d mod %d=%d",a,d,c,ans);
	return 0;
}