杭电HDOJ 1097 解题报告

A hard puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23044    Accepted Submission(s): 8089

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

解题思路：

和1061，我给出的方案1，思路相同。就是找规律，四个一循环，取末位数，进行相乘，然后取最末位数字。

代码：

#include<stdio.h>int main(){    int i,pro;    long a,b;    while(scanf("%ld%ld",&a,&b)!=EOF)    {        pro=1;        b=(b-1)%4;        a=a%10;        for(i=0;i<=b;i++)            pro*=a;        printf("%d\n",pro%10);    }    return 0;}