HDU 1061 Rightmost Digit (求n^n的最低位)

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 52403    Accepted Submission(s): 19868

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

234

 

Sample Output

76
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.
用原始的办法一定会超时！
以下是快速幂的代码：
#include<iostream>#include<stdlib.h>#include<cstdio>#include<algorithm>#include<cstring>#include<queue>#include<math.h>#include<time.h>using namespace std;int PowerMod(int a, int b, int c)//快速幂{    int ans = 1;    a = a % c;    while(b>0)    {        if(b % 2 == 1)        ans = (ans * a) % c;        b = b/2;        a = (a * a) % c;    }    return ans;}int main(){    int t,n;    cin>>t;    while(t--)    {        cin>>n;        cout<<PowerMod(n,n,10)<<endl;    }    return 0;}
以下是找规律的代码：（是比较容易想到的）
int main(){ long long  n; 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}},d,t;//找规律 scanf("%d",&t); while(t--) {  scanf("%lld",&n);  d=n%10;  if(d==0||d==1||d==5||d==6)   printf("%d\n",d);  else if(d==4||d==9)   printf("%d\n",a[d][n%2]);  else   printf("%d\n",a[d][n%4]); } return 0;}