hdu1396--Counting Triangles

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Problem Description

Given an equilateral triangle with n the length of its side, program to count how many triangles in it.

Input

The length n (n <= 500) of the equilateral triangle’s side, one per line.

process to the end of the file

Output

The number of triangles in the equilateral triangle, one per line.

Sample Input

1
2
3

Sample Output

1
5
13

递推题，想法如下：

首先知道了D(1)=1,D(2)=5,D(3)=13;

在手动推出D(4)=27,D(5)=48,然后 不难发现，D(n)=D(n-1)+X,

而这里的X还不确定，根据画出来的图可以看出，当n=4时，D(4)=D(3)+4*2-1+3+2+1+1,

D(5)=D(4)+5*2-1+4+3+2+1+2;

看出规律后，试着假设D(6)=D(5)+6*2-1+5+4+3+2+1+3,

可是画图发现，最后面加的那个数不是3，而是4，即：D(6)=D(5)+6*2-1+5+4+3+2+1+4，

也就是说，D(6)=D(5)+6*2-1+5+4+3+2+1+3+1；

而D(5)=D(4)+5*2-1+4+3+2+1+2+0;

继而再一次假设，当n为偶数时，有D(n)=D(n-1)+n*2-1+(n-1)+(n-2)+…+1+(n-3)+(n-5)+..+1;

当n为奇数时，D(n)=D(n-1)+n*2-1+(n-1)+(n-2)+…+1+(n-3)+(n-5)+..+0;

详细有代码……

代码：

#include <iostream> 
#include <cstdio>
using namespace std;
int ans[510];
int main()
{
    int n, i, j;
    ans[1] = 1;
    for(i = 2; i <= 500; i++)
    {
        ans[i] = ans[i-1] + i*(i+1)/2;
        for (j = i-1; j > 0; j-=2)
        {
            ans[i] += j;
        }
    }
    while(~scanf("%d", &n))
    {
        cout << ans[n] << endl;
    }
}