标签：des blog io os ar strong for div sp
| Memory Limit: 65536K | ||
| Total Submissions: 4637 | Accepted: 1180 |
Description
Given a N × N matrix A, whose element in the i-th row and j-th column Aij 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
12 1 1 2 1 2 2 2 3 2 4 3 1 3 2 3 8 3 9 5 1 5 25 5 10
Sample Output
3 -99993 3 12 100007 -199987 -99993 100019 200013 -399969 400031 -99939
思路:
1 可以看出当j确定的时候i是单调递增的,那么就可以二分得到某个值当j确定时有多少i的值大于它,设为big
2 二分答案当big+ind>n*n(也即全部个数)时,这个值就太小了,增加下界,反之减少上界即可
错误原因 1:全部个数忘了n*n,打成n了 2:上下界错误,看成了1e4 3:读取爆longlong
#include <cstdio>
using namespace std;
long long ind,n;
long long equ(long long i,long long j){
return i*i+j*j+i*j+(i-j)*100000;
}
long long judge(long long mid){
long long big=0;
for(int j=1;j<=n;j++){
int l=0;
int r=n+1;
long long cp;
while(r-l>1){
int m=(r+l)>>1;
cp=equ(m,j);
if(cp==mid){
l=m;
break;
}
else if(cp<mid){
l=m;
}
else{
r=m;
}
}
// printf("binary %I64d col %d %I64d\n",mid,j,l);
big+=n-l;
}
return big;
}
void printe(){
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
printf("%6I64d ",equ(i,j));
}
puts("");
}
}
int main(){
int T;
// freopen("C:\\Users\\Administrator\\Desktop\\input.txt","r",stdin);
scanf("%d",&T);
while(T--){
scanf("%I64d%I64d",&n,&ind);
long long l=-(3*n*n+n*300000),r=-l;
//printe();
while(r-l>1){
long long mid=l+r>>1;
long long big=judge(mid);
if(big>n*n-ind){
l=mid;
}
else {
r=mid;
}
}
printf("%I64d\n",r);
}
return 0;
}
标签：des blog io os ar strong for div sp
原文地址：http://www.cnblogs.com/xuesu/p/3971539.html
