Codeforces Round #569 (Div. 2) A. Alex and a Rhombus

链接:

https://codeforces.com/contest/1180/problem/A

题意:

While playing with geometric figures Alex has accidentally invented a concept of a n-th order rhombus in a cell grid.

A 1-st order rhombus is just a square 1×1 (i.e just a cell).

A n-th order rhombus for all n≥2 one obtains from a n−1-th order rhombus adding all cells which have a common side with it to it (look at the picture to understand it better).

Alex asks you to compute the number of cells in a n-th order rhombus.

思路:

没增加一层,最外层的正方形增加4个。
累加计算。

代码:

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
 
int a[100];
 
int main()
{
    int n;
    cin >> n;
    int res = 1;
    int cnt = 4;
    for (int i = 2;i <= n;i++)
    {
        res += cnt;
        cnt += 4;
    }
    cout << res << endl;
 
    return 0;
}
原文地址:https://www.cnblogs.com/YDDDD/p/11155450.html