POJ 3685

题意：N*N的矩阵，第i行第j列的元素为i*i+j*j+10^5*i-10^5*j+i*j，求第M大的数。

题解：二分枚举答案，算出整个矩阵中比它大的元素个数以及比它小的元素个数（枚举j，二分枚举i）

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LL;
const LL TP=100000,inf=1LL<<40;
LL findmid(LL left,LL right)
{
    if(left+right>=0LL)
        return (left+right)/2;
    else
        return (left+right-1)/2;
}
void update(LL j,LL v,LL n,LL &low,LL &high)
{
    LL left=1,right=n,i,b=j+TP,c=j*j-TP*j-v,a;
    while(left<right)
    {
        i=findmid(left,right);
        a=i*i+b*i+c;
        if(a>=0)right=i;
        else left=i+1;
    }
    a=left*left+left*b+c;
    //printf("left=%lld\nright=%lld\n",left,right);
    if(a==0)
        low+=left-1,high+=n-left;
    else if(a>0)
        low+=left-1,high+=n-left+1;
    else
        low+=n;
}
int cal(LL v,LL n,LL m)
{
    LL low=0,high=0;
    for(LL j=1; j<=n; j++)
        update(j,v,n,low,high);
    //printf("v=%lld %lld %lld\n",v,low,high);
    //system("pause");
    if(m<=low)
        return -1;
    else if(m>n*n-high)
        return 1;
    else
        return 0;
}
int main()
{
    int T;
    for(scanf("%d",&T); T; T--)
    {
        LL n,m,ans,left,right,mid;
        scanf("%lld%lld",&n,&m);
        left=-inf,right=inf;
        while(left<right)
        {
            mid=findmid(left,right);
            //printf("[%lld,%lld]\n",left,right);
            int tp=cal(mid,n,m);
            if(tp==-1)
                right=mid;
            else if(tp==0)
                left=right=mid;
            else
                left=mid+1;
        }
        printf("%lld\n",left);
    }
    return 0;
}