A hard puzzle

最新推荐文章于 2021-04-01 22:58:00 发布

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本文介绍了一种快速计算a^b最后一位数的方法，通过分析不同基数的循环节来简化计算过程，特别适用于大整数的幂运算场景。

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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 66
8 800

Sample Output

9
6

分析：求a^b的最后一位数。找到循环节然后分别对应输出即可。

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

int main(){
    int a,b;
    while(cin>>a>>b){
        a%=10;
        if(a==0||a==1||a==5||a==6)cout<<a<<endl;
        if(a==2){
            if(b%4==0)cout<<6<<endl;
            if(b%4==1)cout<<2<<endl;
            if(b%4==2)cout<<4<<endl;
            if(b%4==3)cout<<8<<endl;
        }
        if(a==3){
            if(b%4==0)cout<<1<<endl;
            if(b%4==1)cout<<3<<endl;
            if(b%4==2)cout<<9<<endl;
            if(b%4==3)cout<<7<<endl;
        }    
        if(a==4){
            if(b%2==0)cout<<6<<endl;
            if(b%2==1)cout<<4<<endl;
        }
        if(a==7){
            if(b%4==0)cout<<1<<endl;
            if(b%4==1)cout<<7<<endl;
            if(b%4==2)cout<<9<<endl;
            if(b%4==3)cout<<3<<endl;
        }
        if(a==8){
            if(b%4==0)cout<<6<<endl;
            if(b%4==1)cout<<8<<endl;
            if(b%4==2)cout<<4<<endl;
            if(b%4==3)cout<<2<<endl;
        }
        if(a==9){
            if(b%2==0)cout<<1<<endl;
            if(b%2==1)cout<<9<<endl;
        }
        
    }
    return 0;
}