HDU 1097 A hard puzzle（找规律，快速幂）

A hard puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35063    Accepted Submission(s): 12599

Problem Description

lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin.
this puzzle describes that: gave a and b,how to know the a^b's the last digit number.But everybody is too lazy to slove this problem,so they remit to you who is wise.

 

Input

There are mutiple test cases. Each test cases consists of two numbers a and b(0<a,b<=2^30)

 

Output

For each test case, you should output the a^b's last digit number.

 

Sample Input

   
   
   
   

    
    
    
    7 66
8 800
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    9
6

   
   
   
   

求a^b.因为只要求得出最后一位，所以只需要进行最后一位的幂运算就行，然而那样对于2^30数据肯定会超时。

这个题很显然是有规律的，例如7,7^1%10=7 , 7^2%10=9 ， 7^3%10=3 ， 7^4%10=1 ， 7^5%10=7

下面是代码

<span style="font-size:18px;">#include <stdio.h>
int a[10];
int main()
{
	int x,y,i;
	while(~scanf("%d %d",&x,&y))
	{
		x=x%10;
		i=0;
		a[0]=x;
		do
		{
			i++;
			a[i]=(a[i-1]*x)%10;
		}while(a[i]!=x);
		
		printf("%d\n",a[(y-1)%i]);
	}
	return 0;
}</span>

刚刚又学了一种解决该题目的方法，叫做快速幂。直接上AC代码，不解释，自行百度吧。

#include <stdio.h>
int PowerMod(int a,int b,int c)
{
	int ans=1;
	a%=c;
	while(b>0)
	{
		if(b%2)
			ans=(ans*a)%c;
		b/=2;
		a=(a*a)%c;		
	}
	return ans;
}

int main()
{
	int a,b;
	while(~scanf("%d %d",&a,&b))
	{
		printf("%d\n",PowerMod(a,b,10));
	}
	
	return 0;
}