poj 3685 二分

Matrix

Time Limit: 6000MS		Memory Limit: 65536K
Total Submissions: 7415		Accepted: 2197

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 

題意：

n*n的矩陣，矩陣中的元素a[i,j]是 i*i+j*j+i*j+1e5*(i-j) ,求矩陣中的第m小的數

代碼：

//首先二分答案mid，而後在矩陣中找小於mid的個數和m比較，i*i+j*j+i*j+1e5*(i-j) 是關於i遞增的即矩陣的每一列都是遞增的，
//因此能夠枚舉列數二分行數來找。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
const ll qq=100000;
ll t,n,m;
ll solve(ll p)
{
    ll sum=0;
    for(ll j=1;j<=n;j++){
        ll l=1,r=n,ans=0;
        while(l<=r){
            ll i=(l+r)>>1;
            if((i*i+j*j+i*j+qq*(i-j))<p) { ans=i;l=i+1; }
            else r=i-1;
        }
        sum+=ans;
    }
    return sum;
}
int main()
{
    scanf("%lld",&t);
    while(t--){
        scanf("%lld%lld",&n,&m);
        ll l=-1e13,r=1e13,ans;
        while(l<=r){
            ll mid=(l+r)>>1;
            ll tmp=solve(mid);
            if(tmp<=m-1) { ans=mid;l=mid+1; }
            else r=mid-1;
        }
        printf("%lld\n",ans);
    }
    return 0;
}