Codeforces 232A - Cycles (构造 + 思维)

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;
}