zoukankan      html  css  js  c++  java
  • HDU_2212

    其实printf输出4个数就行。。

    Problem Description
    A DFS(digital factorial sum) number is found by summing the factorial of every digit of a positive integer. 

    For example ,consider the positive integer 145 = 1!+4!+5!, so it's a DFS number.

    Now you should find out all the DFS numbers in the range of int( [1, 2147483647] ).

    There is no input for this problem. Output all the DFS numbers in increasing order. The first 2 lines of the output are shown below.
     
    Input
    no input
     
    Output
    Output all the DFS number in increasing order.
     
    Sample Output
    1 2 ......
     1 #include <cstdio>
     2 #include <ctime>
     3 int fun(int n)
     4 {
     5    int s=1;
     6    for(int i=2;i<=n;i++)
     7       {
     8          s=s*i;
     9       }
    10    return s;   
    11 }
    12 int main()
    13 {
    14    int s,n,r;
    15    for(int i=1;i<=40585;i++)
    16       {
    17          s=0;n=i;
    18          while(n)
    19             {
    20                r=n%10;
    21                n=n/10;
    22                s=s+fun(r); 
    23             }
    24          if(s==i)
    25             printf("%d\n",i);  
    26       }
    27 }


     

    ——现在的努力是为了小时候吹过的牛B!!
  • 相关阅读:
    emacs jedi
    opencv 基本demo
    emacs列编辑
    observable operator example
    angular keydown 例子
    回调和匿名函数
    gin cors
    angular rxjs
    python dbus note
    视频截图
  • 原文地址:https://www.cnblogs.com/pingge/p/3137979.html
Copyright © 2011-2022 走看看