HDU 1396 Counting Triangles

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递推:a[n] = a[n-1] + n * (n + 1) / 2; 但这并不是结果。

还有一种情况是最底下一行顶角朝下的三角形还没有计算

没有考虑这样情况，我也错了...hehe
观察规律得:left = (n-1) + (n-1)-2 + (n-1)-4... + last(n为奇数, last为2, 反之为1)
所以a[n] = a[n] + left;

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2010-11-22 23:08:04 Accepted 1396 0MS 204K 306 B C Y

 

 

#include <stdio.h>
#define N 501

int main()
{
    int i, n, num[N];;
    
    num[1] = 1;
    for (i = 2; i < N; i++)
    {
        num[i] = num[i-1] + i * (i + 1) / 2; /* 第一种情况 */
        n = i - 1;
        while (n > 0) /* 第二种情况 */
        {
            num[i] += n;
            n -= 2;
        }
    }
    
    while (scanf("%d", &n) != EOF)
    {
        printf("%d\n", num[n]);
    }
}

 

 

 

 

 

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