hdu 1097 A hard puzzle 快速幂取模

最新推荐文章于 2025-09-03 23:14:03 发布

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原文链接：http://www.cnblogs.com/ACFLOOD/p/4402579.html

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#java #php

博客给出ACM题目链接，对题目进行分析，指出是简单的快速幂取模题，因只需输出最后一位，开始可直接mod10，还给出了转载链接。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1097

分析：简单题，快速幂取模，由于只要求输出最后一位，所以开始就可以直接mod10.

/*A hard puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 33036    Accepted Submission(s): 11821


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
*/
//快速幂取模
#include <cstdio>
#include <cstring>
int main()
{
    int a, b;
    while(~scanf("%d%d", &a, &b)){
        a = a % 10;
        int cnt = 1;
        while(b > 0){
            if(b % 2 == 1) cnt = cnt*a%10;
            b = b/2;
            a = a*a%10;
        }
        printf("%d\n", cnt);
    }
    return 0;
}