1600: Big Mod

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本文介绍了一种高效的大数模幂运算算法，适用于处理大数值的B, P, M进行模幂运算。输入包括三个整数B, P, M，输出为计算结果。通过使用位操作和递归调用等技巧提高计算效率。

1600: Big Mod

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	1658	309	Standard

Calculate

for large values of B, P, and M using an efficient algorithm. (That's right, this problem has a time dependency !!!.)

Input

Three integer values (in the order B, P, M) will be read one number per line. B and P are integers in the range 0 to 2147483647 inclusive. M is an integer in the range 1 to 46340 inclusive.

Output

The result of the computation. A single integer.

Sample Input

Sample Output

13
2
13195

#include<iostream>
using namespace std;
int main(){
long long b,p,m,d;
while(cin>>b>>p>>m)
{
     int a[100],k=0;
     if(b==0||m==1){cout<<"0"<<endl;continue; }
     if(p==0){cout<<"1"<<endl;continue;}
     while(p>0){ a[k++]=p%2; p/=2;} k--; d=b;
     int t=1;
     for(int i=0;i<=k;i++)
     {
         if(a[i]) t=(t*d)%m;
         d=(d*d)%m;
     }
     cout<<t<<endl;    
}
return 0;
}

#include<stdio.h>
 long mod(int b,int p,int m)
{
	if(m==1||b==0) return 0;
	if(p==0) return 1;
	if(p%2) return (b%m*mod(b,p-1,m))%m;
	long long temp=mod(b,p/2,m);
	return (temp*temp)%m;
}
int main()
{
	long long b,p,m;
	while(scanf("%d%d%d",&b,&p,&m)!=-1)
	{
		printf("%ld/n",mod(b,p,m));
	}
	return 0;
}