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.
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!!