題干：

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?(x1,?y1)?and?(x2,?y2)?is?.

The pseudo code of the unexpected code is as follows:

input n for i from 1 to ninput 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 nfor j from (i+1) to n++totif (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

題目大意：

給你一個程序，要求讓你出一組數據，使得這組數據會讓程序的tot超過k。換句話說，給出一個最大循環次數k，并且已知如果tot超過k就會TLE。問：能不能給出個樣例，讓這代碼TLE。

解題報告：

? 先找到無論如何都不會讓他TLE的情況，，也就是tot很小，而我們想讓他TLE，，但是死活TLE不了，也就是我們想讓tot盡量大，但是最大情況也是TLE不了的。。。也就是一個break也沒有執行，那么肯定是tot最大的。。。所以我們就是構造一組解，讓他一個break也執行不了，就最會讓他TLE。就行了。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long #define pb push_back #define pm make_pair #define fi first #define se second using namespace std; const int MAX = 2e5 + 5; ll n,k; int main() {cin>>n>>k;if(n * (n-1) / 2 > k) {for(int i = 1; i<=n; i++) {printf("1 %d\n",i);}}else puts("no solution");return 0 ;}

?

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