C. Number of Pairs
You are given an array a of n integers. Find the number of pairs (i,j) (1≤i<j≤n) where the sum of ai+aj is greater than or equal to l and less than or equal to r (that is, l≤ai+aj≤r).

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

i=1 and j=2 (4≤5+1≤7);
i=1 and j=3 (4≤5+2≤7).
Input
The first line contains an integer t (1≤t≤104). Then t test cases follow.

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

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

It is guaranteed that the sum of n overall test cases does not exceed 2?105.

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

Example
inputCopy
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
outputCopy
2
7
0
1
之前沒用過lower_bounder和upper_bounder 做題的時(shí)候沒有想到。二分又懶得寫。。。。。。
所以直接掉大分

#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;} }

哭了。。。。。。

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