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
1 2 3
Sample Output
1 5 13
Author
JIANG, Jiefeng
Source
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;
}