HDU2274-Magic WisKey

Magic WisKey

                                                                            Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                                                         Total Submission(s): 583    Accepted Submission(s): 329

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

293326663922174245386 

Sample Output

000380771921238

Source

HDU 8th Programming Contest Site（1）

 

 题意： 选定五个数字，他们形成一个五位数N。告诉你这五个数字全排列后组成的 120 个五位数的和减去其中一个五位数的结果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。

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <algorithm>#include <cmath>#include <queue>#include <stack>#include <vector>#include <set>#include <map>using namespace std;int n;int main(){    int a[100005];    for(int i=0;i<=99999;i++)    {        int sum=0;        int n=i;        while(n>0)        {            sum+=n%10;            n/=10;        }        int ans=sum*24*11111;        a[i]=ans-i;    }    while(~scanf("%d",&n))    {        for(int i=0;i<99999;i++)        {            if(a[i]==n)                printf("%05d\n",i);        }    }    return 0;}