【light-oj】-’1307 - Counting Triangles（二分）

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

 

题意：n个木棍，问有多少可以构成三角形。

题解：两两结合，判断第三条边可以的个数。（个数二分计算即可）

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define LL long long
LL a[2010];
int main()
{
int u,ca=1;
scanf("%d",&u);
while(u--)
{
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%lld",&a[i]);
sort(a+1,a+1+n);
int sum,ans=0;
for(int i=1;i<n;i++)
{
for(int j=i+1;j<n;j++)
{
sum=a[i]+a[j];
int pos=lower_bound(a+j,a+1+n,sum)-a;
ans+=(pos-j-1);
}
}
printf("Case %d: %d\n",ca++,ans);
}
return 0;
}