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 ......各个位上阶乘的和 是最大是9*9！这样来求解。999999999 9个9 最大的各个位上的阶乘的和为3265920=9*9！
只需要遍历到上面这个数就可以了，不用遍历到21亿，否则会超时
#include<stdio.h> int a[11]; int f(int n) { int i,sum=1; for(i=1;i<=n;i++) sum*=i; return sum; } int main() { a[0]=1; int i,num,sum,p; for(i=1;i<=9;i++) a[i]=f(i); for(i=1;i<=300500;i++) { num=i;sum=0; while(num) { p=num%10; num/=10; sum+=a[p]; } if(i==sum) printf("%d\n",i); } return 0; }