HDU1097 A hard puzzle
来源:互联网 发布:太祖书法 知乎 编辑:程序博客网 时间:2024/06/04 01:09
A hard puzzle
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 29415 Accepted Submission(s): 10581
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.
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
7 668 800
Sample Output
96
#include <stdio.h>#include <string.h>const char *sam[] = {"0", "1", "2486", "3971", "46", "5", "6", "7931", "8426", "91"};int main(){ int a, b; while(scanf("%d%d", &a, &b) == 2) printf("%c\n", sam[a % 10][(b-1) % strlen(sam[a%10])]); return 0;}
0 0
- hdu1097 A hard puzzle
- A hard puzzle(HDU1097)
- hdu1097 A hard puzzle
- HDU1097 A hard puzzle
- HDU1097 A hard puzzle
- HDU1097 A hard puzzle
- hdu1097 A hard puzzle
- hdu1097 A hard puzzle
- hdu1097 A hard puzzle
- hdu1097 A hard puzzle
- hdu1097 - A hard puzzle
- hdu1097 A hard puzzle
- HDU1097:A hard puzzle(快速幂求模)
- hdu1097 A hard puzzle(快速幂)
- ACM-数论之A hard puzzle——hdu1097
- HDU1097 A hard puzzle O(1)算法实现
- hdu1097 A hard puzzle hdu5500 Reorder the Books
- A hard puzzle
- Source Insight设置
- Android复习练习四(QQ登陆界面回显用户名密码,使用/data/data/包名/路径 )
- UITableview的的秒用
- Nightmare (HDU 1072)—— BFS
- C++中使用MediaInfo库获取视频信息
- HDU1097 A hard puzzle
- UVa10669 - Three powers
- javascript基础知识总结
- 浅谈Entity Framework中的数据加载方式
- TopCoder-AvoidRoads
- NOJ 2005 BDD和CSS的旅游 状压DP
- Matlab plot、fplot、ezplot三者区别
- jQuery Easy UI整理笔记目录
- Lighttpd博客收藏