hdu-1012-u Calculate and (题)

1 #include <iostream> 
 2  using  namespace std;
 3  int main() {
 4      int jiecheng[ 10 ];
 5      jiecheng[ 0 ] = jiecheng[ 1 ] = 1 ;
 6      for ( int i= 2 ; i< 10 ; i++ ) {
 7          jiecheng[i] = jiecheng[i- 1 ] * i;
 8      }
 9      printf( " ne\n " );
 10      printf("- -----------\n");
11     printf("0 1\n");
12     printf("1 2\n");
13     printf("2 2.5\n");
14     double num[10];
15     num[2] = 2.5;
16     for (int i=3; i<10; i++) {
17         num[i] = 1.0/jiecheng[i] + num[i-1];
18         printf("%d %.9lf\n", i,num[i]);
19     }
20     return 0;
21 }