杭电acm1097 A hard puzzle
来源:互联网 发布:专业音乐录音软件 编辑:程序博客网 时间:2024/05/20 10:21
杭电acm1097
A hard puzzle
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18812 Accepted Submission(s): 6707
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
7 668 800
Sample Output
96
Author
eddy
Recommend
JGShining
思想:
尾数为0~9的数字的n次方都可以看做是四个数的循环,如:a为一的尾数,都是1。。a的最后一位为2时,当n次方时,始终都是2,4,8,6的循环,其他同理,所以可以
把b求余4来算,但防止b除以4后,余数为零,所以至少保证有a的4次方存在。。。
实现代码如下:
#include <iostream>
using namespace std;
int main()
{
int a,b,i,y;
while(cin>>a>>b)
{
if(a<0 || b<0)
break;
a%=10;
b%=4;
y = (a*a*a*a);//防止b求余后为0;没进入for循环,至少保证怎样都有值
for(i=0;i<b;i++)
{
y*=a;
y%=10;
}
cout<<y%10<<endl;
}
return 0;
}
- 杭电acm1097 A hard puzzle
- 杭电1097 A hard puzzle
- 杭电 HDU 1097 A hard puzzle
- 【杭电】[1097]A hard puzzle
- [水题]杭电1097 A hard puzzle
- 杭电acm1097
- HDOJ 1097 A hard puzzle 杭电 ACM
- 杭电1097-A hard puzzle(快速幂)
- 杭电1097A hard puzzle(快速幂求余)
- 杭电OJ(HDOJ)1097题:A hard puzzle(数论)
- 杭电OJ 1097 A hard puzzle(我最后的博客水题报告)
- A hard puzzle
- 1097 A hard puzzle
- hdu1097 A hard puzzle
- 1097A hard puzzle
- 1097 A hard puzzle
- HDU - A hard puzzle
- A hard puzzle(HDU1097)
- post和get提交中文数据的转码问题
- Java 双向链表
- 杭电acm1046 Gridland
- 卷动列 && 子窗口颜色
- 键盘驱动
- 杭电acm1097 A hard puzzle
- 采集点建立后生成更新时提示:文档ID:3 模板文件不存在,无法解析文档
- 杭电acm 1040 As Easy As A+B
- 共阴和共阳数码管编码表
- POJ 1125 Stockbroker Grapevine
- 发一批 phpcmsV9 系统漏洞mysql
- [每日学习笔记][2012.08.02]使用Java理解程序逻辑(十)
- PHPcmsV9任意文件读取漏洞拿Shell
- 本原勾股数组