HDU 1097 a hard puzzle

LIcker丶D

于 2016-03-12 20:23:55 发布

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分类专栏： C/C++---OJ题解析

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本文介绍了一个高效的方法来计算两个整数a和b的幂次运算的最后一位数字，通过观察规律和使用取模操作简化计算。

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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

Sample Output

  
  
   
   9

   
   6
   
   

   
   简单来说，就是求前一个数的最后一个数的b次方的问题，很容易就看出来有规律，四个一循环，所以很容易就解决了。
   
   #include<stdio.h>
#include<math.h>
int main()
{
	int a,b,a1,b1,sum;
	while(scanf("%d%d",&a,&b)!=EOF)
	{
		a1=a%10;
		b1=b%4;
		if(b1==0){
			sum=((int)pow(a1,4))%10;
	     	printf("%d\n",sum);
		}
		else{
			sum=((int)pow(a1,b1))%10;
	     	printf("%d\n",sum);
		} 
	}
 }