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

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.

用原始的辦法一定會超時！

以下是快速冪的代碼：

#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)//快速冪 {     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; }

以下是找規律的代碼：（是比較容易想到的）

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;//找規律  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; }