POJ 3685 Matrix 二分 函数单调性 难度:2

 

	Memory Limit: 65536K
Total Submissions: 4637		Accepted: 1180

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

思路:
1 可以看出当j确定的时候i是单调递增的,那么就可以二分得到某个值当j确定时有多少i的值大于它,设为big
2 二分答案当big+ind>n*n(也即全部个数)时,这个值就太小了,增加下界,反之减少上界即可
错误原因 1:全部个数忘了n*n,打成n了 2:上下界错误,看成了1e4 3:读取爆longlong

#include <cstdio>

using namespace std;



long long ind,n;

long long equ(long long i,long long j){

    return i*i+j*j+i*j+(i-j)*100000;

}

long long judge(long long mid){

    long long big=0;

    for(int j=1;j<=n;j++){

        int l=0;

        int r=n+1;

        long long cp;

        while(r-l>1){

            int m=(r+l)>>1;

            cp=equ(m,j);

            if(cp==mid){

                l=m;

                break;

            }

            else if(cp<mid){

                l=m;

            }

            else{

                r=m;

            }

        }

 //       printf("binary %I64d col %d %I64d\n",mid,j,l);

        big+=n-l;

    }

    return big;

}

void printe(){

    for(int i=1;i<=n;i++){

        for(int j=1;j<=n;j++){

            printf("%6I64d ",equ(i,j));

        }

        puts("");

    }

}

int main(){

    int T;

   // freopen("C:\\Users\\Administrator\\Desktop\\input.txt","r",stdin);

    scanf("%d",&T);

    while(T--){

        scanf("%I64d%I64d",&n,&ind);

        long long l=-(3*n*n+n*300000),r=-l;

        //printe();

        while(r-l>1){

            long long mid=l+r>>1;

            long long big=judge(mid);

            if(big>n*n-ind){

                l=mid;

            }

            else {

                r=mid;

            }

        }

        printf("%I64d\n",r);

    }

    return  0;

}