HDU 1061 Rightmost Digit



Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 36563    Accepted Submission(s): 13871

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.
 

 

Author

Ignatius.L

 

求N的N次方的最后一位数。

快速幂。。

假如8^4。可以看成64^2   。。

这样运算次数减少了三次。

假如8^5 我们可以先将一个8存起来，变成8^4。。

然后一直反复操作。。

代码

#include <stdio.h>#include <cmath>#include <vector>#include <map>#include <time.h>#include <cstring>#include <set>#include<iostream>#include <queue>#include <stack>#include <algorithm>using namespace std;#define inf 0x6f6f6f6f#define mod 10long long wei(long long k){    long long n=k;    long long r=1;    while(k)    {        if(k&1)  //k次幂为奇数，可以存起一个数。          r=r*n%mod;        n=n*n%mod;        k>>=1;   //位运算，加快速度。    }    return r;}int main(){    long long k;    int t;    scanf("%d",&t);    while(t--)    {        cin>>k;        long long ans=wei(k);        cout<<ans<<endl;    }    return 0;}