超大数次幂再求余

H - Super A^B mod C

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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Description

Given A,B,C, You should quickly calculate the result of A^B mod C. 

(1<=A,C<=1000000000,1<=B<=10^1000000).

Input

There are multiply testcases. Each testcase, there is one line contains 

three integers A, B and C, separated by a single space.

Output

For each testcase, output an integer, denotes the result of A^B mod C.

Sample Input

3 2 4
2 10 1000

Sample Output

1
24

某大伽的超级代码

简便代码：

//模拟乘法。将乘法编程二进制加法。

#include 
unsigned __int64 mul (unsigned __int64 a,unsigned __int64 b,unsigned __int64 c)
{
 	unsigned __int64 res=0,tmp=a%c;
	while(b)
 	{
		if(b&1 && (res+=tmp) >= c) 
			res -= c;
		if((tmp <<= 1) >= c) 
			tmp -= c;
		b>>=1;
 	}
	return res;
}

int main(void)
{
	 unsigned __int64 a,b,c,y;
	 while (scanf("%I64u %I64u %I64u",&a,&b,&c)!=EOF)
	 {
		y = 1;
		while(b)
 		{

			if(b&1) y = mul (y,a,c);
			a=mul(a,a,c);
			b>>=1;
 		}
		printf("%I64u\n",y%c);
 	}
	return 0;
}