POJ-3685---Matrix (二分）

Matrix

Time Limit: 6000MS Memory Limit: 65536KTotal Submissions: 7248 Accepted: 2144

Description

Given a N × N matrix A, whose element in the i-th row andj-th columnA_ij is an number that equals i² + 100000 ×i +j² - 100000 × j + i × j, you are to find theM-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

121 12 12 22 32 43 13 23 83 95 15 255 10

Sample Output

3-99993312100007-199987-99993100019200013-399969400031-99939

Source

POJ Founder Monthly Contest – 2008.08.31, windy7926778

题意：有一个N*N的矩阵，矩阵中每个点的值由题中给的公式计算出，问矩阵中第m小的数是什么；

思路：首先可以打个表会发现矩阵的每一列是随i增大单调递增的，那么现在就可以用二分来求答案，对于每次二分得到的mid ，再去矩阵中二分每一列求得小于mid的数的个数，最后求得答案；注意这题要用long long，开始wa了一发，然后全部改成long long过了；（附上一张打的表）

AC代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long LL;LL n,m;LL calculate(LL i,LL j)//计算矩阵中的值{    return i*i+100000*i+j*j-100000*j+i*j;}LL Count(LL val)//统计比val小的数的个数{    LL cnt=0;    for(LL i=1;i<=n;i++)//枚举每一列    {        LL l=0,r=n+1;        while(r-l>1)        {            LL mid=(l+r)>>1;            if(calculate(mid,i)>=val) r=mid;            else l=mid;        }        cnt+=l;    }    return cnt;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%lld%lld",&n,&m);        LL l=-100000*n,r=n*n+100000*n+n*n*2;//最初r和l取极大极小值        while(r-l>1)        {            LL mid=(l+r)>>1;            if(Count(mid)>=m) r=mid;//            else l=mid;        }        printf("%lld\n",l);    }    return 0;}