快速幂求a^b结果的最后一位

最新推荐文章于 2021-11-21 14:38:00 发布

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最新推荐文章于 2021-11-21 14:38:00 发布

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本文链接：https://blog.youkuaiyun.com/csdn16890/article/details/76782774

本文介绍了一种解决特定数学问题的方法：通过快速幂取模算法来找出两个数a和b相乘后的结果的最后一位数字。该算法适用于多个测试案例，并通过样例输入输出展示了算法的具体应用。

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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

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int k_mi(int a, int b) { //快速幂取模
int ans = 1,t = a % 10;
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;
}