nyoj473 A^B Problem (快速幂)
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A^B Problem
时间限制:1000 ms | 内存限制:65535 KB
难度:2
- 描述
- Give you two numbers a and b,how to know the a^b's the last digit number.It looks so easy,but everybody is too lazy to slove this problem,so they remit to you who is wise.
- 输入
- There are mutiple test cases. Each test cases consists of two numbers a and b(0<=a,b<2^30)
- 输出
- For each test case, you should output the a^b's last digit number.
- 样例输入
7 668 800
- 样例输出
96
- 提示
- There is no such case in which a = 0 && b = 0。
- 来源
- hdu
- 上传者
- ACM_丁国强
#include <stdio.h>int main(){int a,b,_a,s;while(scanf("%d %d",&a,&b)!=EOF){if(a==0&&b==0)break;s=1;while(b){if(s>=10)s=s%10;if(a>=10)a=a%10;if(b%2==1)s=s*a;a=a*a;b=b/2;}if(s>=10)s=s%10;printf("%d\n",s);}return 0;}
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