HDU 1061-Rightmost Digit

题目

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 67183    Accepted Submission(s): 25198

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.

题目链接：​​http://acm.hdu.edu.cn/showproblem.php?pid=1061​​

思路

题意大概就是 求 

 的最后一位数。

有两个思路，一个是找规律，另一个是用快速幂

找规律：

0的n次方尾数还是0，1的n次方尾数还是1，5的n次方尾数还是5，6的n次方同样是6。

2的n次方位数为{6，2，4，8}以此类推。

因为尾数是固定的，所以说0~9都是可以推出来的。

然后根据指数n，取余，得到结果。

快速幂：

快速幂这就不用说了，老江湖了。

根据下边的这个公式，我们只用算最后一位数的n次方即可。

AC代码在下方。

代码

快速幂解决代码

#include<cstdio>

using namespace std;
const int MOD = 10;
#define mod(x) ((x)%MOD)

int poww(int a,int b){
    int res = 1,base=a;
    while(b){
        if(b&1)res = mod(res*base);
        base = mod(base*base);
        b >>= 1;
    }
    return res;
}

int main(){
    int t,n;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        printf("%d\n",poww(mod(n),n));
    }
    return 0;
}

找规律

#include<cstdio>

using namespace std;
const int MOD = 10;
#define mod(x) ((x)%MOD)

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}};

int main(){
    int t,n;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);

        int t = mod(n);

        if(t == 0||t == 1||t == 5||t == 6){

        }else if(t == 4||t == 9){
            t = a[t][n%2];
        }else{
            t = a[t][n%4];
        }
        printf("%d\n",t);
    }
    return 0;
}