HDU 2274 Magic WisKey

Posted 2020-08-13 风中的簌雨

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：

 1 #include<stdio.h>  
 2 #include<string.h>  
 3 int a[6],s[100005],sum;  
 4 int main()  
 5 {  
 6     for(int i=0;i<=99999;i++) {  //先打表  
 7         int t=i,k=0,sum;  
 8         while(t>0) {  
 9             a[k++]=t%10;  
10             t/=10;  
11         }  
12         int tmp=0;  
13         for(int j=0;j<k;j++)  
14             tmp+=a[j];  
15         s[i]=tmp*24*11111-i;  
16     }   
17     while(~scanf("%d",&sum)) {  //直接查询  
18         for(int i=0;i<=99999;i++)  
19             if(sum==s[i]) printf("%05d\n",i);  
20     }  
21     return 0;  
22 }