NYOJ779 兰州烧饼

原题链接

需要考虑特殊情况：n<=k时，用时为2.

#include <stdio.h>

int main(){
int n, k, t, s;
while(scanf("%d%d", &n, &k) == 2){
if(n <= k) s = 2;
else if(2*n % k == 0) s = 2*n / k;
else s = 2*n / k + 1;
printf("%d\n", s);
}
return 0;
}

800423

长木

兰州烧饼

Accepted

8

232

C/C++

04-07 14:24:06