首页 > 代码库 > nyoj473 A^B Problem (高速幂)
nyoj473 A^B Problem (高速幂)
A^B Problem
时间限制:1000 ms | 内存限制:65535 KB
难度:2
- 描写叙述
- 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.
- 例子输入
-
7 66 8 800
- 例子输出
-
9 6
- 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; }
nyoj473 A^B Problem (高速幂)
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。