HDU 1061.Rightmost Digit-规律题 or 快速幂取模

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 55522    Accepted Submission(s): 20987

Problem Description

Given a positive integer N, you should output the most right digit of N^N.

 

Input

The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains a single positive integer N(1<=N<=1,000,000,000).

 

Output

For each test case, you should output the rightmost digit of N^N.

 

Sample Input

2

3

4

 

Sample Output

7

6

Hint

In the first case, 3 * 3 * 3 = 27, so the rightmost digit is 7. In the second case, 4 * 4 * 4 * 4 = 256, so the rightmost digit is 6.

 

 

第一个，找规律，代码:

#include<stdio.h>
typedef long long ll;
int main(){
    int a[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}};
    ll m;
    int n,i,h;
    while(~scanf("%d",&n)){
        for(int i=0;i<n;i++){
            scanf("%lld",&m);
            h=m%10;
            if(h==0||h==1||h==5||h==6)
                printf("%d",h);
            else if(h==4||h==9)
               printf("%d",a[h][m%2]);
            else
                printf("%d",a[h][m%4]);
                printf("
");
            }
    }
   return 0;
}

第二个，快速幂取模，代码:

#include<stdio.h>
typedef long long ll;
ll mod=1e5;
ll pow(ll a,ll b){
    ll ans=1;while(b!=0){
        if(b%2==1)
            ans=ans*a%mod;
            a=a*a%mod;
            b=b/2;
    }
    return ans;
}
int main(){
    ll n,m,ans;
    while(~scanf("%lld",&n)){
        while(n--){
            scanf("%lld",&m);
            ans=pow(m,m)%10;
            printf("%lld
",ans);
        }
    }
    return 0;
}