POJ 3685 二分

Posted 2020-09-29 LuZhiyuan

Matrix

Time Limit: 6000MS		Memory Limit: 65536K
Total Submissions: 7034		Accepted: 2071

Description

Given a N × N matrix A, whose element in the i-th row and j-th column A_ij is an number that equals i² + 100000 × i + j² - 100000 × j + i × j, you are to find the M-th smallest element in the matrix.

Input

The first line of input is the number of test case.
For each test case there is only one line contains two integers, N(1 ≤ N ≤ 50,000) and M(1 ≤ M ≤ N × N). There is a blank line before each test case.

Output

For each test case output the answer on a single line.

Sample Input

12

1 1

2 1

2 2

2 3

2 4

3 1

3 2

3 8

3 9

5 1

5 25

5 10

Sample Output

3
-99993
3
12
100007
-199987
-99993
100019
200013
-399969
400031
-99939

Source

POJ Founder Monthly Contest – 2008.08.31, windy7926778 

题意：

有n*n的矩阵，A(ij)的值是 i² + 100000 × i + j² - 100000 × j + i × j，求矩阵中第k小的数。

输入 t

输入t行n k

输出结果

代码：

//和上一题差不多，当j固定时表达式随i的增大而增大，先二分第k小数的值s，然后枚举j，
//找小于等于s的i的个数有几个，用这个值与k比较来二分。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
int t;
ll n,k;
ll make(ll i,ll j){
    return i*i+j*j+i*100000-j*100000+i*j;
}
bool solve(ll m){
    ll cnt=0;
    for(int j=1;j<=n;j++){
        int l=1,r=n,ans=0;
        while(l<=r){
            int i=(l+r)>>1;
            if(make(i,j)<=m){
                ans=i;
                l=i+1;
            }else r=i-1;
        }
        cnt+=ans;
    }
    return cnt>=k;
}
int main()
{
    scanf("%d",&t);
    while(t--){
        scanf("%lld%lld",&n,&k);
        ll l=-100000*n;
        ll r=n*n+n*n+100000*n+n*n,ans;
        while(l<=r){
            ll m=(l+r)>>1;
            if(solve(m)){
                ans=m;
                r=m-1;
            }else l=m+1;
        }
        printf("%lld\n",ans);
    }
    return 0;
}