Light OJ：1307 Counting Triangles（二分+暴力枚举）

1307 - Counting Triangles

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Time Limit: 2 second(s)Memory Limit: 32 MB

You are given N sticks having distinct lengths; you have to form some triangles using the sticks. A triangle is valid if its area is positive. Your task is to find the number of ways you can form a valid triangle using the sticks.

Input

Input starts with an integer T (≤ 10), denoting the number of test cases.

Each case starts with a line containing an integer N (3 ≤ N ≤ 2000). The next line contains N integers denoting the lengths of the sticks. You can assume that the lengths are distinct and each length lies in the range[1, 10⁹].

Output

For each case, print the case number and the total number of ways a valid triangle can be formed.

Sample Input

Output for Sample Input

3

5

3 12 5 4 9

6

1 2 3 4 5 6

4

100 211 212 121

Case 1: 3

Case 2: 7

Case 3: 4

 

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PROBLEM SETTER: JANE ALAM JAN

题目大意：给了你n个边长，保证每个边长都是唯一的，输出可组成多少种三角形。

解题思路：普通三重for循环肯定超时。这里我们将边长排序，两个for循环控制两个最小边，然后二分计算后面有多少条边小于这两个边即可。

代码如下：

#include <cstdio>#include <algorithm>using namespace std;int n;int tmp;int a[2010];int main(){int t;scanf("%d",&t);int kcase=1;while(t--){scanf("%d",&n);for(int i=0;i<n;i++){scanf("%d",&a[i]);}sort(a,a+n);//将边排序 int cnt=0;for(int i=0;i<n;i++){for(int j=i+1;j<n;j++){tmp=a[i]+a[j];//枚举两个短边之和 int pos=lower_bound(a+j,a+n,tmp)-a;//找他大于的第三边的下标 if(pos==j)//找不到不加 {continue;}cnt=cnt+(pos-j-1);//找到的话，算出个数，加上去 }}printf("Case %d: %d\n",kcase++,cnt);}return 0;}