【LightOJ】1307 - Counting Triangles（二分）

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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

问能组成的三角形数。用二分查找计数就行了，数据不大。

代码如下：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <map>
#include <vector>
#include <stack>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(a,b) memset(a,b,sizeof(a))
#define LL long long
#define PI acos(-1.0)
int main()
{
int u;
int n;
int Case = 1;
int num[2016];
scanf ("%d",&u);
while (u--)
{
scanf ("%d",&n);
for (int i = 1 ; i <= n ; i++)
scanf ("%d",&num[i]);
printf ("Case %d: ",Case++);
sort(num+1 , num+1+n);
int ans = 0;
for (int i = 1 ; i < n ; i++)
{
for (int j = i + 1 ; j <= n ; j++)
ans += lower_bound(num+1+j , num+1+n , num[i]+num[j]) - 1 - num - j;
}
printf ("%d\n",ans);
}
return 0;
}