hdu2212 DFS

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......
#include<stdio.h>int f[15],an;int pass(unsigned int sum)//判断每一位的介层和是不是等于本身，反回是几位数{    unsigned int su=0,a=sum;    int k,t=0;    while(a>0)    {        k=a%10;        su+=f[k];        a/=10;        t++;    }    return (su==sum)?t:0;}void DFS(unsigned int sum,int k,int t)//k是sum要达到多少位数，t是当前的数是几位数{    int e;    if(pass(sum)==k)//只有当反回的位数等要求的位数才输出，不然会重复    printf("%d\n",sum);    if(t<k)//当前的位数达不到要求的位数才继续加    for(e=0;e<10;e++)    if((e!=0||t!=0)&&sum*10+e<=2147483647)//(e!=0||t!=0)是控制最高位不为0    {        DFS(sum*10+e,k,t+1);    }}int main(){    int i;    f[0]=1;    for(i=1;i<10;i++)    f[i]=f[i-1]*i;    for(i=1;i<6;i++)    DFS(0,i,0);}