HDU 2274 Magic WisKey

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2274

Magic WisKey

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 568    Accepted Submission(s):
323

Problem Description

On New Year Festival, Liu Qian’s magic impressed on
little WisKey’s heart and he wants to learn some magic to make himself
stronger.
  One day, he met a cowman named LinLe. Linle is very nice, he told
little WisKey the mysteries of magic. Now, little WisKey began to perform the
magic to you.
  “Hello, Everybody. I have five decimal numbers named a, b, c,
d, e, (0 <= a, b, c, d, e <= 9) and I rearranged them, and then combined
them into a number, for example <a, b, c, d, e> = a*10000 + b*1000 + c*100
+ d*10 + e*1. You know the number of permutations is 5! = 120. So you have 120
numbers in your hands, you can pick a number N from the 120 numbers and
calculate the sum S of remain 119 numbers. AHA~, If you tell me the S, I can
guess the N~!”
  It’s easy? Okay, you can challenge this.

 

Input

Each line will contain an integer S. Process to end of
file.

 

Output

For each case, output all possible N, print the number
N with 5 digits, including the leading zeros, one line per case.
I promise
every case have one solution at least. If have many N, please output them from
small to large in one line, separate them with a blank space.

 

Sample Input

2933266

6392217

4245386

 

Sample Output

00038

07719

21238

 

题目大意：选定五个数字，他们形成一个五位数N。你事先不知道是哪五个数字，但你知道这五个数字全排列后组成的 120 个五位数的和减去该五位数的结果S，求N。

 

解题思路：五位数总共的可能就是从00000~99999。所以先打表暴力枚举求出所有五位数的S，然后对于每个输入的S，查找对应的N。

求五位数全排列之和的方法：假设五个数字分别是a,b,c,d,e。首先考虑a，a在万位上会出现4！次，千位上也会出现4！次，百位十位个位同理，所以a出现的每个地方总和为a*4！*（10000+1000+100+10+1）。b,c,d,e与a类似。所以总和为(a+b+c+d+e)*4!*11111。

AC代码：

 #include<stdio.h>
 #include<string.h>
 int a[],s[],sum;
 int main()
 {
     for(int i=;i<=;i++) {  //先打表
         int t=i,k=,sum;
         while(t>) {
             a[k++]=t%;
             t/=;
         }
         int tmp=;
         for(int j=;j<k;j++)
             tmp+=a[j];
         s[i]=tmp**-i;
     }
     while(~scanf("%d",&sum)) {  //直接查询
         for(int i=;i<=;i++)
             if(sum==s[i]) printf("%05d\n",i);
     }
     return ;
 }