AcWing 1227. Divide chocolate (y total two classic two-division templates)

The difference between the
two templates for topic y : see whether mid is placed in l or r

int bsearch_1(int l, int r)
{
    
    
    while (l < r)
    {
    
    
        int mid = l + r >> 1;
        if (check(mid)) r = mid;    // check()判断mid是否满足性质
        else l = mid + 1;
    }
    return l;
}

int bsearch_2(int l, int r)
{
    
    
    while (l < r)
    {
    
    
        int mid = l + r + 1 >> 1;
        if (check(mid)) l = mid;
        else r = mid - 1;
    }
    return l;
}

Idea: Use dichotomy to find the most suitable number of divisions. If there is more chocolate, change mid to l, and if less, change mid to r-1.

If you find that the answer is 1 more than the actual one when doing a dichotomous question, you can directly subtract 1 when outputting the answer

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 100010;

int n, k;
int h[N], w[N];

bool check(int mid)
{
    
    
    int res = 0;
    for (int i = 0; i < n; i ++ )
    {
    
    
        res += (h[i] / mid) * (w[i] / mid);//计算可以分给多少个人巧克力 
        if (res >= k) return true;
    }

    return false;
}

int main()
{
    
    
    scanf("%d%d", &n, &k);
    for (int i = 0; i < n; i ++ ) scanf("%d%d", &h[i], &w[i]);

    int l = 1, r = 1e5;
    while (l < r)//y总的第二经典模板 
    {
    
    
        int mid = l + r +1 >> 1;
        if (check(mid)) l = mid;
        else r = mid -1;
    }

    printf("%d\n", r);

    return 0;
}