1001. 害死人不偿命的(3n+1)猜想 (15)
较简单,直接代码实现:
#include <cstdio> int main() { int n; scanf("%d",&n); int k =0; while(n!=1) { if(n%2 == 0) { k++; n = n/2; } else { n = 3*n+1; } } printf("%d\n",k); return 0; }
来源:https://www.cnblogs.com/yanyun888/p/6103233.html