lower_bound( )和upper_bound( ) Cf Round #725 (Div. 3) C

lower_bound会找出序列中第一个大于等于x的数
upper_bound会找出序列中第一个大于x的数

C. Number of Pairs

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an array aa of nn integers. Find the number of pairs (i,j)(i,j) (1≤i<j≤n1≤i<j≤n) where the sum of ai+ajai+aj is greater than or equal to ll and less than or equal to rr (that is, l≤ai+aj≤rl≤ai+aj≤r).

For example, if n=3n=3, a=[5,1,2]a=[5,1,2], l=4l=4 and r=7r=7, then two pairs are suitable:

• i=1i=1 and j=2j=2 (4≤5+1≤74≤5+1≤7);
• i=1i=1 and j=3j=3 (4≤5+2≤74≤5+2≤7).

Input

The first line contains an integer tt (1≤t≤1041≤t≤104). Then tt test cases follow.

The first line of each test case contains three integers n,l,rn,l,r (1≤n≤2⋅1051≤n≤2⋅105, 1≤l≤r≤1091≤l≤r≤109) — the length of the array and the limits on the sum in the pair.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109).

It is guaranteed that the sum of nn overall test cases does not exceed 2⋅1052⋅105.

Output

For each test case, output a single integer — the number of index pairs (i,j)(i,j) (i<ji<j), such that l≤ai+aj≤rl≤ai+aj≤r.

输入

4
3 4 7
5 1 2
5 5 8
5 1 2 4 3
4 100 1000
1 1 1 1
5 9 13
2 5 5 1 1

输出

2
7
0
1

lower_bound( begin,end,num)：从数组的begin位置到end-1位置二分查找第一个大于或等于num的数字，找到返回该数字的地址，不存在则返回end。通过返回的地址减去起始地址begin,得到找到数字在数组中的下标。

upper_bound( begin,end,num)：从数组的begin位置到end-1位置二分查找第一个大于num的数字，找到返回该数字的地址，不存在则返回end。通过返回的地址减去起始地址begin,得到找到数字在数组中的下标。

#include <iostream>
#include <algorithm>
using namespace std;
#define int long long
int ch[205001];
signed main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int n, l, r;
        cin >> n >> l >> r;
        for (int i = 0; i < n; i++)
        {
            cin >> ch[i];
        }
        sort(ch, ch + n);
        int sum = 0;
        for (int i = 0; i < n; i++)
        {
            int a = lower_bound(ch + i + 1, ch + n, l - ch[i]) - ch;
            int b = upper_bound(ch + i + 1, ch + n, r - ch[i]) - ch;
            int c = b - a;
            sum += c;
        }
        cout << sum << endl;
    }
}