Codeforces 232A

题目链接：

232A - Cycles(点击打开)

题意：

要构成一个存在 (k) 个三元环的图，需要多少个点，输出顶点数 (n)，并输出图。

题解：

题目中的任何图都可以用 (90)~ (100)个顶点构造完成。

Proof that (100) vertices are always enough for the given restrictions on (n).

- For some (p) after first (p) iterations we will have a complete graph of (p) vertices.

- Now we have exactly (C(p, 3)) triangles. Consider (p) such that (C(p, 3)  ≤ k) and (C(p, 3)) is maximal.

- For the given restrictions (p ≤ 85).

- From this moment, if we add (u) from some vertex, we increase the total number of 3-cycles on (C(u, 2).)

- So we have to present a small number that is less than (C(85, 3)) as sum of (C(i, 2)).

The first number we subtruct will differ (C(85, 1)) on some value not greater than (C(85, 1) = 85), because (C(n, k) - C(n - 1, k) = C(n - 1, k - 1)).

- The second number we subtruct will differ the number we have on some value not greater than (C(14, 1) = 14.)

- and so on.

- For every (k) it's enough to use not more that (90) vertices.

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int g[123][123];
int main()
{
	int k;
	cin>>k;//要求的三元环个数 
	for(int i=0;i<90;i++)//依次增加顶点 
	{
		int sum = 0;//三元环个数 
		for(int j = 0; j < i && sum <= k;j++)
		{
			k-=sum;
			sum++;
			g[i][j] = g[j][i] = 1;
		}
	 }
	cout<<90<<endl;
	for(int i=0;i<90;i++)
	{
		for(int j=0;j<90;j++)
		{
			int ans = g[i][j] ? 1:0;
			cout<<ans;
		}
		cout<<endl;
	}
	return 0;
}