题目很简单，就是我们在写二分的时候，会遇到这样的情况：

mid =left +(right-left)/2;

        if (nums == k)return mid;
        if (nums>k)left = mid;
        if (nums<k)right = mid-1;

若我们debug会发现其可能会一致卡在left = mid，例如（0，1），那么就会一直循环

其实可以改为

mid =left +(right-left)/2+1;

那么就可以了。

#include <cstdio>
#include <cmath>
const int N =1e3+10;
int n,k;
int q[N];

int binary_search(int left,int right){
    int mid;
    int nums=0;
    while (left<right){
        nums = 0;
        mid =left +(right-left)/2+1;
        for (int i = 0;i<n;i++){
            nums+=q[i]/mid;
        }
        if (nums == k)return mid;
        if (nums>k)left = mid;
        if (nums<k)right = mid-1;
        printf("%d %d\n",left,right);
    }
    return left;
}
int main(){
    scanf("%d%d",&n,&k);
    int max_=0;
    for (int i=0;i<n;i++){
        scanf("%d",&q[i]);
        if (q[i]>max_)max_ = q[i];
    }

    printf("%d",binary_search(0,max_));
    return 0;


}