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): 1160    Accepted 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

123

 

Sample Output

1513

 

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;elset3=(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;}