UVA10162 Last Digit【数学规律】

Give you a integer number N (1 ≤ n ≤ 2 ∗ 10¹⁰⁰). Please compute
S = 1¹ + 2² + 3³ + . . . + N^N
Give the last digit of S to me.
Input
Input file consists of several N’s, each N a line. It is ended with N = 0.
Output
For each N give a line containing only one digit, which is the last digit of S.
Sample Input
1
2
3
0
Sample Output
1
5
2

问题链接：UVA10162 Last Digit
问题简述：给定n，计算S = 1¹ + 2² + 3³ + . . . + N^N的最后1位数字。
问题分析：数学规律题，通过打表找规律，可以发现到1-100后开始循环。进一步可以发现每20个又有规律。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* UVA10162 Last Digit */

#include <bits/stdc++.h>

using namespace std;

const int m[] = {0,1,5,2,8,3,9,2,8,7,7,8,4,7,3,8,4,1,5,4};
const int N = 100 + 2;
char s[N];

int main()
{
    while(~scanf("%s", s)) {
        int len = strlen(s);
        if(len == 1 && s[0] == '0') break;

        int a = s[len - 1] - '0';
        if(len > 1)
            a += (s[len - 2] - '0') * 10;

        printf("%d\n", (m[a % 20] + a / 20 * 4) % 10);
    }

    return 0;
}