HDU1097暴力打表找规律a^b

A hard puzzle

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 44659 Accepted Submission(s): 16333

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.

Output
For each test case, you should output the a^b’s last digit number.

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

思路

1、还是打表找规律，找a^b值的个位数

打表找规律代码

打表a从1-15，b从1-10找规律

#include <stdio.h>#include <iostream>using namespace std;int main(){    for(int i=1;i<=15;i++)    {        long int sum=1;        for(int j=1;j<=10;j++)        {            sum*=i;            if(sum>100)                sum%=100;            cout<<sum%10<<"  ";        }        cout<<endl;    }    return 0;}

可以看出a从1-10顺序规律，b也可以看出每项规律。放入二维数组即可

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代码

#include <iostream>#include <math.h>#include <stdio.h>using namespace std;int main(){    long int a,b;    int s[10][5]={{0},{1},{2,4,8,6},{3,9,7,1},{4,6},{5},{6},{7,9,3,1},{8,4,2,6},{9,1}};    while(cin>>a>>b)    {        a=a%10;        if(a==0)            cout<<s[0][0]<<endl;        if(a==1)            cout<<s[1][0]<<endl;        if(a==2)        {            int m=(b%4)-1;            if(m==-1)                m=3;            cout<<s[2][m]<<endl;        }        if(a==3)        {            int m=(b%4)-1;            if(m==-1)                m=3;            cout<<s[3][m]<<endl;        }        if(a==4)        {            int m=(b%2)-1;            if(m==-1)                m=1;            cout<<s[4][m]<<endl;        }        if(a==5)        {            cout<<s[5][0]<<endl;        }        if(a==6)            cout<<s[6][0]<<endl;        if(a==7)        {            int m=(b%4)-1;            if(m==-1)                m=3;            cout<<s[7][m]<<endl;        }        if(a==8)        {            int m=(b%4)-1;            if(m==-1)                m=3;            cout<<s[8][m]<<endl;        }        if(a==9)        {            int m=(b%2)-1;            if(m==-1)                m=1;            cout<<s[9][m]<<endl;        }    }    return 0;}