CodeForces - 312C The Closest Pair Co 構造
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x
The pseudo code of the unexpected code is as follows:
input n for i from 1 to n input the i-th point's coordinates into p[i] sort array p[] by increasing of x coordinate first and increasing of y coordinate second d=INF //here INF is a number big enough tot=0 for i from 1 to n for j from (i+1) to n ++tot if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be //out of the loop "for j" d=min(d,distance(p[i],p[j])) output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
- All the points must be distinct.
- |xi|, |yi| ≤ 109.
- After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0 0 1 1 0 1 1
Input
2 100
Output
no solution
題解:說的是先按x再按y排序之後,跑完迴圈,讓tot超過k,操蛋啊,剛開始構造的x遞減,就納悶為啥一直錯,靠,
還排序。。。。
排序的話,直接讓所有x相等就行了,y不一樣就行。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e5+10;
int n,k;
int main()
{
while(~scanf("%d%d",&n,&k))
{
if((n-1)*n/2<=k) printf("no solution\n");
else
{
for(int i=1;i<=n;i++)
printf("%d %d\n",0,i);
}
}
return 0;
}