poj 3685 Matrix (二分+枚举+二分)

Matrix

Time Limit: 6000MS Memory Limit: 65536KTotal Submissions: 4233 Accepted: 1035

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 

思路：

1.二分答案，根据矩阵中小于这个数的个数与m的值比较来二分。

2.怎样在矩阵中查找呢？方法也是枚举+二分。仔细观察会发现矩阵每列的值满足单调性（每行不满足），所以想到枚举每列，在每列中找到该列小于这个数的值求和就够了。

ps：二分时注意向上取整还是向下取整，然后注意用long long 就够了。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>//#pragma comment (linker,"/STACK:102400000,102400000")#define maxn 50005#define mod 1000000000#define INF 0x3f3f3f3fusing namespace std;typedef long long ll;ll n,m,ans;ll f(ll x,ll y){    return x*x+100000*x+y*y-100000*y+x*y;}ll calnum(ll k){    ll i,j;    ll le,ri,mid,sum=0;    for(j=1;j<=n;j++)  // 对每列进行枚举    {        le=1;        ri=n+1;        while(le<ri)   // 找比k小的数        {            mid=(le+ri)>>1;            if(f(mid,j)>=k) ri=mid;            else le=mid+1;        }        sum+=le-1;    }    return sum;}void solve(){    ll le,mid,ri,t;    ri=1LL<<50;    le=-ri;    while(le<ri)  // 二分找答案    {        mid=(le+ri)>>1;        t=calnum(mid);        if(t>=m) ri=mid;        else le=mid+1;    }    ans=le-1;}int main(){    int i,j,t;    scanf("%d",&t);    while(t--)    {        scanf("%lld%lld",&n,&m);        solve();        printf("%lld\n",ans);    }    return 0;}