HDU/HDOJ 1396 浙大月赛2003年

Counting Triangles

Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1160Accepted Submission(s): 573

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

Author

JIANG, Jiefeng

Source

ZOJ Monthly, June 2003

我是这样思考的。

可以看出来，边长+1意味着三角形多了一行

那么可以针对多的那一行做特殊计算，因为上一次边长的三角形是没有变化的！

经过非常穷凶极恶的公式推导终于得到了最后的答案

我的代码：

#include<stdio.h>

int ans[505];

int cal(int n)
{
	int t1,t2,t3;
	t1=2*n-1;
	t2=(n-1)*n/2;
	if(n&1)
		t3=(n-1)*(n-3)/4;
	else
		t3=(n-2)*(n-2)/4;
	return t1+t2+t3;
}

void init()
{
	int i;
	ans[1]=1;
	for(i=2;i<=500;i++)
		ans[i]=ans[i-1]+cal(i);
}

int main()
{
	int n;
	init();
	while(scanf("%d",&n)!=EOF)
	{
		printf("%d\n",ans[n]);
	}
	return 0;
}