poj 3685 矩阵问题 查找第K小的值

题意：N阶矩阵Aij= i2 + 100000 × i + j2 – 100000 × j + i × j，求第M小的元素。

思路：双重二分

1. 考虑到，aij是跟着i递增的，所以i可以作为一个二分搜索
2. 统计比 mid小的个数 
3. 如果个数小于 m的话 mid太小 
4. 反之 mid太大

解决问题的代码：

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <math.h>
using namespace std;

typedef long long ll;

ll t, n, m;
ll f(ll i, ll j)
{
    return i * i + 100000 * i + j * j - 100000 * j + i * j;
}
bool solve(ll mid)
{
    ll less = 0;
    for (int j = 1; j <= n; j++)
    {
        int lb = 0, ub = n + 1;
        while (ub - lb > 1)
        {
            int i = (ub + lb) / 2;
            if (f(i, j) <mid)
            {
                lb = i;
            }
            else ub = i;
        }
        less += lb;
    }
    return less < m;
}
int main()
{
    scanf("%lld", &t);
    while (t--)
    {
        scanf("%lld%lld", &n, &m);
        ll lb = -100000 * n;
        ll ub = n * n + 100000 * n + n * n + n * n;
        while (ub - lb > 1)
        {
            ll mid = (ub + lb) / 2;
            if (solve(mid)) lb = mid;
            else ub = mid;
        }
        printf("%lld\n", lb);
    }
    return 0;
}