POJ

二分kth，答案满足的条件为：m ≤ 小于等于x的值数cnt_x。x和cntx单调不减，随着x增大，条件成立可表示为：0001111。

本地打一个小型的表可以发现列编号j固定时候，目标函数f(i,j)似乎具有单调性。

变形，f(i,j) = (i+100000+j)*i + j² - 100000，可以看出确实具有单调性。

于是得到如下算法： 二分x，统计cnt时候，枚举j，再套一个二分。

/*********************************************************
*            ------------------                          *
*   author AbyssalFish                                   *
**********************************************************/
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<queue>
#include<vector>
#include<stack>
#include<vector>
#include<map>
#include<set>
#include<algorithm>
#include<cmath>
#include<numeric>
using namespace std;

typedef long long ll;
ll n, m;


const int d = 1e5;

inline ll f(ll j,ll i){ return (i+d+j)*i + (j-d)*j; }

int upp_b(ll j, ll x)
{
    if(f(j,1) > x) return 0;
    int lb = 1, ub = n, md;
    while(lb < ub ){
        md = (lb+ub+1)>>1;
        //f(j,md)>x? ub = md+1:lb = md+1;
        f(j,md)<=x? lb = md:ub = md-1;
    }
    return lb;
}

ll solve()
{
    if(m == 1) {
        ll ans = (3*n+d)*n;
        for(ll j = 1; j <= n; j++){
            ans = min(ans, (1+d+j)+(j-d)*j);
        }
        return ans;
    }
    if(m == n*n){
        ll ans = -(3*n+d)*n;
        for(ll j = 1; j <= n; j++){
            ans = max(ans, (n+d+j)*n-d*j+j*j);
        }
        return ans;
    }
    ll lb = -(3*n+d)*n, ub = -lb, md;
    while(lb < ub){
        md = (lb+ub)>>1;
        ll upb = 0;
        for(ll j = 1; j <= n; j++){
            upb += upp_b(j,md);
        }
        upb >= m?ub = md:lb = md+1;
    }
    return lb;
}

//#define LOCAL
int main()
{
#ifdef LOCAL
    freopen("in.txt","r",stdin);
#endif
    int T; scanf("%d",&T);
    while(T--){
        scanf("%I64d%I64d",&n,&m);
        printf("%I64d
", solve());
    }
    return 0;
}