二分板子

求最小值

求最小值

int binary()
{
    int l = 0, r = ll, mid;
    while(l < r)
      {
        mid = (l + r) >> 1;
        if(check(mid)) r = mid;  //大多数题只要改改check()即可
        else l = mid + 1;
      }
    return l;
}

求最大值

求最大值

int binary()
{
    int l = 0, r = ll, mid;
    while(l < r)
      {
        mid = (l + r + 1) >> 1;
        if(check(mid)) r = mid - 1;
        else l = mid;
      }
    return l;
}

面对整数时近似万能的二分

万能二分

int binary(int n)
{
    int l = 1, r = maxn, ans = 0;
    while(l <= r)
      {
        int mid = (l + r) >> 1;
        if(c[mid] > a[n]) ans = mid, l = mid + 1;  //判断条件与ans记录位置因题而异
        else r = mid - 1;
      }
    return ans;
}

转自：https://blog.csdn.net/Mashiro_ylb/article/details/78469151