1307-counting Triangles

1307-counting Triangles

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Time Limit:2 second (s)

Memory limit:32 MB

You is given N sticks having distinct lengths; You have a to form some triangles using the sticks. A Triangle is valid if it 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 is 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 the lengths is 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 is 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

Problem Setter:jane ALAM Jan Test Instructions: How many triangles can be formed on your side. Idea: First violent combination of two side, then two points to query the third edge can be.

1#include <stdio.h>2#include <algorithm>3#include <iostream>4#include <string.h>5#include <queue>6#include <stdlib.h>7#include <math.h>8#include <stack>9#include <vector>Ten#include <map> One using namespacestd; AtypedefLong LongLL; - intans[2005]; -typedefstructpp the { -      Short intx; -      Short inty; - } SS; +SS ak[4000005]; - intMainvoid) + { A     inti,j,k; atscanf"%d",&k); -     ints; -      for(s=1; s<=k; s++) -     { -         intn,m; -scanf"%d",&n); in          for(i=0; i<n; i++) -         { toscanf"%d",&ans[i]); +         } -Sort (ans,ans+n); the         intCnt=0; *          for(i=0; i<n; i++) $         {Panax Notoginseng              for(j=i+1; j<n; J + +) -             { theak[cnt].x=i; +ak[cnt].y=J; Acnt++; the             } +         } -LL sum=0; $          for(i=0; i<cnt; i++) $         { -             intL=0; -             intr=n-1; the             intmaxx=Max (ans[ak[i].x],ans[ak[i].y]); -             intminn=min (ans[ak[i].x],ans[ak[i].y]);Wuyi             intId=0; theL=0; -r=n-1;intid1=-1; Wu              while(l<=R) -             { About                 intMid= (L+R)/2; $                 if(ans[mid]+minn>Maxx) -                 { -id1=mid; -r=mid-1; A                 } +                 ElseL=mid+1; the             } -L=0; $r=n-1;intid2=-1; the              while(l<=R) the             { the                 intMid= (L+R)/2; the                 if(ans[mid]<maxx+Minn) -                 { inId2=mid; theL=mid+1; the                 } About                 Elser=mid-1; the             } the             if(id1!=id2) the             { +sum+=id2-id1+1; -                 if(ak[i].x>=id1&&ak[i].x<=id2) thesum--;Bayi                 if(ak[i].y<=id2&&ak[i].y>=id1) thesum--; the             } -         } -printf"Case %d:", s); theprintf"%lld\n", sum/3); the     } the     return 0; the}

1307-counting Triangles