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 DescriptionGiven a positive integer N, you should output the most right digit of N^N. InputThe 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). OutputFor each test case, you should output the rightmost digit of N^N. Sample Input234 Sample Output76

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[][]={{},{},{,,,},{,,,},{,},{},{},{,,,},{,,,},{,}};
    ll m;
    int n,i,h;
    while(~scanf("%d",&n)){
        for(int i=;i<n;i++){
            scanf("%lld",&m);
            h=m%;
            if(h==||h==||h==||h==)
                printf("%d",h);
            else if(h==||h==)
               printf("%d",a[h][m%]);
            else
                printf("%d",a[h][m%]);
                printf("\n");
            }
    }
   return ;
}

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=;while(b!=){
        if(b%==)
            ans=ans*a%mod;
            a=a*a%mod;
            b=b/;
    }
    return ans;
}
int main(){
    ll n,m,ans;
    while(~scanf("%lld",&n)){
        while(n--){
            scanf("%lld",&m);
            ans=pow(m,m)%;
            printf("%lld\n",ans);
        }
    }
    return ;
}