PAT_1001(中文)

 1 #include<stdio.h>
 2 
 3 int main()
 4 {
 5     int n;
 6     int step = 0;
 7     scanf("%d", &n);
 8     while( 1 != n )
 9     {
10         if( (n%2) == 0 )
11         {
12             n = n / 2;    
13             ++step;
14         }    
15         else
16         {
17             n = 3 * n + 1;
18             n = n / 2;    
19             ++step;
20         }
21     }
22     printf("%d\n", step);
23     return 0;
24 }

 

posted @ 2013-04-15 21:38  多解方程式  阅读(128)  评论(0编辑  收藏  举报