POJ3685-Matrix

Matrix

Time Limit: 6000MS Memory Limit: 65536KTotal Submissions: 6722 Accepted: 1958

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

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：n阶矩阵Aij= i*i + 100000 × i + j*j – 100000 × j + i × j，求第m小的元素

解题思路：我先打n为比较小的表，发现很有规律，大小规律是从右上往左下依次增大，但是wa了，据说这个规律到n为5*10^4就不一定保持了。所以这题应该二分第m小的数，又因为 i*i + 100000 × i + j*j – 100000 × j + i × j是关于i递增的，再次二分每一列，找到比它小的数

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <queue>#include <cmath>#include <map>#include <bitset>#include <set>#include <vector>#include <functional>using namespace std;#define LL long longconst LL INF=0x3f3f3f3f3f3f3f3f;LL m,n;bool check(LL x){    LL sum=0;    for(LL i=1;i<=n;i++)    {        LL l=1,r=n,ans=0;        while(l<=r)        {            LL mid=(l+r)/2;            if(mid*mid+100000*mid+i*i-100000*i+mid*i<=x) {ans=mid,l=mid+1;}            else r=mid-1;        }        sum+=ans;    }    if(sum>=m) return 1;    else return 0;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%lld%lld",&n,&m);        LL l=-INF,r=INF,ans;        while(l<=r)        {            LL mid=(l+r)/2;            if(check(mid)) {ans=mid;r=mid-1;}            else l=mid+1;        }        printf("%lld\n",ans);    }    return 0;}