hdu Counting Triangles

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

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        int d[5];
        d[0]=0;
        d[1]=1;
        d[2]=5;
        d[3]=13;
        if(n<=3)
        {
            printf("%d\n",d[n]);
            continue;
        }
        int i;
        for(i=4;i<=n;i++)
        {
            d[4]=d[3]+2*i-1+(i*i-i)/2;
            if(i%2==0)
                d[4]=d[4]+(i-3+1)*(i-2)/4;
            else
                d[4]=d[4]+(i-3)*(i-1)/4;
            d[3]=d[4];
        }
        printf("%d\n",d[4]);
    }
    return 0;
}


        

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……