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扩展题
AcWing 789. 数的范围
该题会同时用到两种模板
C++ 代码
#include <iostream>
using namespace std;
typedef long long LL;
const int N = 100010;
int n, m;
int h[N], w[N];
bool check(int mid)
{
LL res = 0;//res为可以切出来的矩形的个数
for (int i = 0; i < n; i ++ )
{
res += (LL)h[i] / mid * (w[i] / mid);
if (res >= m) return true;
}
return false;
}
int main()
{
scanf("%d%d", &n, &m);//读入n和m
for (int i = 0; i < n; i ++ ) scanf("%d%d", &h[i], &w[i]);//读入矩形的长和宽
int l = 1, r = 1e5;
while (l < r)
{
int mid = l + r + 1 >> 1;//(l+r)/2,因为+的优先级高于>>,所以不用加括号
if (check(mid)) l = mid;
else r = mid - 1;
//因为后两行这样写,所以应该是上取整,将mid = l + r >> 1改为mid = l + r + 1 >> 1
}
printf("%d\n", r);
return 0;
}