思想:
1e5
。pair<int, int> a[N]
存储边长数据,第 a[i]
块巧克力可分割的数量为 (a[i].first / t) * (a[i].second / t)
。代码:
#include <bits/stdc++.h>
using namespace std;
int n, m;
const int N = 1e6 + 3;
pair<int, int> a[N];
bool check(int t){ //判断是否满足条件
int cnt = 0;
for(int i = 0; i < n; i ++){
cnt += (a[i].first / t) * (a[i].second / t);
if(cnt >= m) return 1; //当前分割数量已经满足,提前返回 true
}
return 0;
}
void solve(){
cin >> n >> m;
for(int i = 0; i < n; i ++) cin >> a[i].first >> a[i].second;
int l = 1, r = 1e5;
while(l < r){
int mid = l + r + 1 >> 1;
if(check(mid)) l = mid;
else r = mid - 1;
}
cout << l << endl; //二分终点即为答案
}
int main(){
solve();
return 0;
}