HDU--1097A hard puzzle（快速幂）

最新推荐文章于 2019-05-01 19:06:49 发布

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本文介绍了一种使用快速幂算法解决特定数学问题的方法：给定两个整数a和b，如何高效地计算a^b的结果的最后一位数字。通过提供C语言实现的示例代码，展示了算法的具体步骤。

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A hard puzzle

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 44416 Accepted Submission(s): 16199

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
Author
eddy 这道题用快速幂作。 #include<stdio.h>
int P(int a,int b)
{
	int t=1;
	a%=10;
	while(b!=0){
		if(b&1){
			t=t*a%10;
		}	
		b/=2;
	    a=a*a%10;		
	}
	return t;
}
int main()
{
	int a,b;
	while(scanf("%d %d",&a,&b)!=EOF){
		printf("%d\n",P(a,b));
	}
return 0;
}