題意：N*N的矩陣，第i行第j列的元素爲i*i+j*j+10^5*i-10^5*j+i*j，求第M大的數。
題解：二分枚舉答案，算出整個矩陣中比它大的元素個數以及比它小的元素個數（枚舉j，二分枚舉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;
}