51Nod 1016 水仙花数 V2(组合数学,枚举打表法)

1016 水仙花数 V2

基准时间限制：1 秒 空间限制：131072 KB 分值: 160        

难度：6级算法题
               

水仙花数是指一个 n 位数 ( n≥3 )，它的每个位上的数字的 n 次幂之和等于它本身。（例如：1^3 + 5^3 + 3^3 = 153，1634 = 1^4 + 6^4 + 3^4 + 4^4）。

给出一个整数M，求 >= M的最小的水仙花数。

 

Input

一个整数M（10 <= M <= 10^60）

Output

输出>= M的最小的水仙花数，如果没有符合条件的水仙花数，则输出：No Solution

Input示例

300

Output示例

370
题目链接：http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1016
分析：

一道賊变态的题，如果按常规出牌，绝对做不出来，必然会超时，或者说，我肯定是做不成的。
但是如果投机取巧，那么这就是一道很简单的题，虽然它数据范围高达60位，但是水仙花数却是有限的，只有89个，所以，我们完全可以打表做题。那么剩下的问题就出现了，这些水仙花数是啥？

想知道这些数都是啥，可以选择两种手段，第一暴力解题，获得数据，但是我想这也忒复杂了，虽然是暴力解题，但是依然存在很多问题，就算你写出来代码，想获得所有的水仙花数依然需要很长很长时间等待哦，毕竟运算量惊人。
于是我只好选择第二种办法了，找度娘喽……

这是水仙花数……

看着好爽啊，这么长……

接下来要考虑的问题就是比大小，这也好解决，位数不同的比位数，相同的再逐位比大小。大概就是这个样子吧。剩下的就没有什么难点了。

无耻的打表徒，来一发AC：

 #include <stdio.h>
 #include <string.h>

 int compare(char *a, char *b, int len)
 {
     for (int i = ; i < len; i++)
     {
         if (b[i] > a[i])
         {
             return ;
         }
         else if (b[i] < a[i])
         {
             return ;
         }
     }
     return ;
 }

 int main(int argc, const char * argv[])
 {
     char NarNum[][] = {"","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","","",""};
     int NarNumLen;
     char num[];
     scanf("%s", num);
     int NumLen = (int)strlen(num);
     for (int i = ; i < ; i++)
     {
         NarNumLen = (int)strlen(NarNum[i]);
         if ((NumLen == NarNumLen && compare(num, NarNum[i], NumLen)) || NumLen < NarNumLen)
         {
             printf("%s\n", NarNum[i]);
             return ;
         }
     }

     puts("No Solution");
     return ;
 }