二分求解 三角形 stl的应用 涉及范围的二分查找可以先求上界再算下界，结果即上界减下界

 二分

Time Limit:2000MS      Memory Limit:32768KB      64bit IO Format:%lld & %llu

 

Description

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

3

5

3 12 5 4 9

6

1 2 3 4 5 6

4

100 211 212 121

Sample Output

Case 1: 3

Case 2: 7

Case 3: 4

 

 

 1 #include <iostream>

 2 #include <stdio.h>

 3 #include <algorithm>

 4 #include <math.h>

 5 

 6 using namespace std;

 7 typedef long long LL;

 8 LL a[2000+5];

 9 int n;

10 

11 int Binary_search(int x)

12 {

13     int l = 0,r = n-1;

14     while(l<=r)

15     {

16         int mid = (l+r)/2;

17         if(a[mid] == x)

18             return mid;

19         else if(a[mid] > x)

20             r= mid-1;

21         else

22             l = mid +1;

23     }

24     return l;

25 

26 }

27 

28 int main()

29 {

30     int T;

31     cin>>T;

32     int flag = 0;

33     while(T--)

34     {

35 

36         cin>>n;

37         for(int i = 0; i < n; i++)

38             scanf("%lld",&a[i]);

39         sort(a,a+n);

40         int ans = 0;

41         for(int i = 0; i < n-1; i++)

42             for(int j = i+1; j < n; j++)

43             {

44                 LL sum = a[i] + a[j];

45                 LL cha = abs(a[j] - a[i]);

46 

47                 int up = Binary_search(sum)-1;

48                 if(up <= j)

49                     continue;

50                 int low = Binary_search(cha);

51                 if(low < j)

52                     low = j;

53                 ans +=  (up-low);

54 

55 

56             }

57 

58         cout<<"Case "<<++flag<<": "<<ans<<endl;

59 

60     }

61     return 0;

62 }

View Code

 

 1 #include <iostream>

 2 #include <stdio.h>

 3 #include <algorithm>

 4 #include <math.h>

 5 

 6 using namespace std;

 7 typedef long long LL;

 8 LL a[2000+5];

 9 int n;

10 

11 int Binary_search(int x)

12 {

13     int l = 1,r = n;

14     while(l<r)

15     {

16         int mid = (l+r)/2;

17         if(a[mid] == x)

18             return mid;

19         else if(a[mid] > x)

20             r= mid;

21         else

22             l = mid +1;

23     }

24     return l;

25 

26 }

27 

28 int main()

29 {

30     int T;

31     cin>>T;

32     int flag = 0;

33     while(T--)

34     {

35 

36         cin>>n;

37         for(int i = 0; i < n; i++)

38             scanf("%lld",&a[i]);

39         sort(a,a+n);

40         int ans = 0;

41         for(int i = 0; i < n-1; i++)

42             for(int j = i+1; j < n; j++)

43             {

44                 LL sum = a[i] + a[j];

45                 LL cha = abs(a[j] - a[i]);

46 

47                 int up = Binary_search(sum)-1;

48                 if(up <= j)

49                     continue;

50                 int low = Binary_search(cha);

51                 if(low < j)

52                     low = j;

53                 ans +=  (up-low);

54 

55 

56             }

57 

58         cout<<"Case "<<++flag<<": "<<ans<<endl;

59 

60     }

61     return 0;

62 }

View Code

 1 #include <iostream>

 2 #include <stdio.h>

 3 #include <algorithm>

 4 #include <math.h>

 5 

 6 using namespace std;

 7 typedef long long LL;

 8 LL a[2000+5];

 9 int n;

10 

11 int main()

12 {

13     int T;

14     cin>>T;

15     int flag = 0;

16     while(T--)

17     {

18 

19         cin>>n;

20         for(int i = 0; i < n; i++)

21             scanf("%lld",&a[i]);

22         sort(a,a+n);

23         int ans = 0;

24         for(int i = 0; i < n-1; i++)

25             for(int j = i+1; j < n; j++)

26             {

27                 LL sum = a[i] + a[j];

28                 LL cha = abs(a[j] - a[i]);

29 

30                 int up = upper_bound(a+j+1,a+n,sum)-a;

31 

32 

33                 int low = lower_bound(a+j+1,a+n,cha)-a;

34 

35                 ans +=  (up-low);

36 

37 

38             }

39 

40         cout<<"Case "<<++flag<<": "<<ans<<endl;

41 

42     }

43     return 0;

44 }

View Code