HDU 1097.A hard puzzle【快速幂或规律】【8月12】

A hard puzzle

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

 

题目意思：求a^b的最后一位数，问题转化成a^b%10,代码如下：

#include<cstdio>int mod(int a,int b,int k){    int ans=1;    a=a%k;    while(b>0){        if(b%2==1) ans=(ans*a)%k;        b/=2;        a=(a*a)%k;    }    return ans;}int main(){    int a,b;    while(scanf("%d%d",&a,&b)==2)        printf("%d\n",mod(a,b,10));    return 0;}

这是直接的求法。看别人的题解，竟然还有一个规律：

尾数为0,1,5,6的不管是多少次方尾数依然不变，而尾数为4和9的每2次循环，

2,3,7,8为每4次循环。循环结果如下：

0，1，5，6：位数永远是0，1，5，6

2：6，2，4，8循环

3：1，3，9，7循环

4：6，4循环

7：1，7，9，3循环

8：6，8，4，2循环

9：1，9循环

好的，可以水过了：

#include<cstdio>int div[10]={1,1,4,4,2,1,1,4,4,2};int f[10][4]={{0},{1},{6,2,4,8},{1,3,9,7},{6,4},{5},{6},{1,7,9,3},{6,8,4,2},{1,9}};int main(){    int a,b;    while(scanf("%d%d",&a,&b)==2)        printf("%d\n",f[a%10][b%div[a%10]]);    return 0;}