标签：acm hdu algorithm 快速幂 模
A hard puzzle
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32633 Accepted Submission(s): 11672
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
Author
eddy
AC代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int k_mi(int a, int b) { //快速幂取模
int t = a % 10, ans = 1;
while(b) {
if(b&1) ans = (ans * t) % 10;
t = (t * t) % 10;
b >>= 1;
}
return ans % 10;
}
int main() {
int a, b;
while(scanf("%d %d", &a, &b) != EOF) {
printf("%d\n", k_mi(a, b));
}
return 0;
}
HDU - 1097 - A hard puzzle （快速幂取模）
标签：acm hdu algorithm 快速幂 模
原文地址：http://blog.csdn.net/u014355480/article/details/44247119
