FZU 1752 A^B mod C

 Problem 1752 A^B mod C

Accept: 840    Submit: 3625

Time Limit: 1000 mSec    Memory Limit : 32768 KB

 Problem Description

Given A,B,C, You should quickly calculate the result of A^B mod C. (1<=A,B,C<2^63).

 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 42 10 1000

 Sample Output

124

 Source

FZU 2009 Summer Training IV--Number Theory

大数处理。数据范围达到2^63，所以要用unsigned long long 。

但还是要爆，所以采用加法，对于耗时间的%运算，直接利用减法替代。

#include<stdio.h>
#include<string.h>
#include<math.h>
#define LL unsigned __int64
LL quickmod(LL a,LL b,LL c)
{
	LL ans=0;
	a=a%c;
	while(b)
	{
	   if(b&1)ans+=a;
	   if(ans>=c)ans-=c;
	   a=a+a;
	   if(a>=c)a=a-c;
	   b=b/2;
	}
	return ans;
}
int main()
{
	LL a,b,c,i,j,ans;
	while(scanf("%I64u%I64u%I64u",&a,&b,&c)!=EOF)
	{
		ans=1;
		a=a%c;
		while(b)
		{
			if(b%2)ans=quickmod(ans,a,c);
			a=quickmod(a,a,c);
			b=b/2;
		}
		printf("%I64u\n",ans);
	}
	return 0;
}

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