[C] Beer Land

Beer Land

Time Limit: 1 Sec Memory Limit: 64 MB

Description

King ACM 2002 received some clear ground and some money through inheritance. Terms of the will and testament require the heir to build n towns. He was also required to build the least quantity of two-way roads between the towns such that one could travel from any town to any other town even if one road was closed under repair. Carefully, all such roads must be constructed as geometrically straight lines. Further, the condition that a traveller be able to get from one town to another (even with one road closed) needs to be satisfied by traversing the roads as straight lines in their entirety — not stopping partway along one road to switch to another. Because of rich soil in those lands, the future inhabitants of the place have decided to produce beer. On the crossroads (a crossroad is merey the intersection of two or more roads) the king plans to build beer stands (one stand per a crossroad). So, if three or four (or twenty) roads intersect in a single point, there’s still only one appropriate place for a beer stand. The King must choose locations for the towns and the roads. You need to advise him so that he builds as many beer stands as possible.

Input

The first line of input contains the number of test cases. For each case there is a line containing a single integer n (1 ≤ n ≤ 32767). This represents the number of towns to be built.

Output

For each n, print a line containing the maximum number of beer stands possible to build.

Sample Input

3
3
4
5

Sample Output

0
1
5

[做本题体会]

我认为本题的突破点为找到其中所蕴含的规律。（找到点数与交点之间的规律）
下图为我找规律的分析过程（啊！字有点丑，哈哈哈～）：


找到规律后，代码的编写就变得简单了。
以下为我的代码：

#include <stdio.h>

int main() {
//   测试组数
    int n_group;
    scanf("%d",&n_group);
    int count,test;
    for (count=1; count<=n_group;count++) {
        scanf("%d",&test);
        if(test<=3)
        {
            printf("%d\n",0);
        }
        if (test>3) {
        if (test%2!=0) {
            printf("%d\n",test*(test-3)/2);
        }
        else
        {
            printf("%d\n",test*(test-4)/2+1);
//          printf("%d\n",(2*(test-3)+(test-4)*(test-2))/2);
        }
        }
    }
    return 0;
}