hdoj-2212DFS

DFS

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9257    Accepted Submission(s): 5648

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

12......

 

题目链接

这题结果就4个数，循环不用到2147483647，否则根本停不下= 。=

#include<cstdio>#include<cstring>#include<iostream>using namespace std;int a[15];int jc(int n){    int i;    int sum=1;    for(i=1; i<=n; i++ )        sum*=i;    return sum;}int main(){    int i;    for(i=0;i<=9;i++)    {        a[i]=jc(i);    }    for(i=1;i<=350005;i++)    {        int num=i;        int sum=0;        int k;        while(num)        {            k=num%10;            num/=10;            sum+=a[k];        }        if(sum==i)            printf("%d\n",i);    }    return 0;}