# 計算給定總和的對 > 原文： [https://www.geeksforgeeks.org/count-pairs-with-given-sum/](https://www.geeksforgeeks.org/count-pairs-with-given-sum/) 給定一個整數數組和一個數字“和”，找到該數組中總和等于“和”的整數對的數量。 **示例**： ``` Input : arr[] = {1, 5, 7, -1}, sum = 6 Output : 2 Pairs with sum 6 are (1, 5) and (7, -1) Input : arr[] = {1, 5, 7, -1, 5}, sum = 6 Output : 3 Pairs with sum 6 are (1, 5), (7, -1) & (1, 5) Input : arr[] = {1, 1, 1, 1}, sum = 2 Output : 6 There are 3! pairs with sum 2. Input : arr[] = {10, 12, 10, 15, -1, 7, 6, 5, 4, 2, 1, 1, 1}, sum = 11 Output : 9 ``` 預期時間復雜度`O(n)` 一個**簡單解決方案**遍歷每個元素，并檢查數組中是否有另一個數字可以添加到該數字以求和。 ## C++ ```cpp // C++ implementation of simple method to find count of // pairs with given sum. #include <bits/stdc++.h> using namespace std; // Returns number of pairs in arr[0..n-1] with sum equal // to 'sum' int getPairsCount(int arr[], int n, int sum) { ????int count = 0; // Initialize result ????// Consider all possible pairs and check their sums ????for (int i=0; i<n; i++) ????????for (int j=i+1; j<n; j++) ????????????if (arr[i]+arr[j] == sum) ????????????????count++; ????return count; } // Driver function to test the above function int main() { ????int arr[] = {1, 5, 7, -1, 5} ; ????int n = sizeof(arr)/sizeof(arr[0]); ????int sum = 6; ????cout << "Count of pairs is "? ?????????<< getPairsCount(arr, n, sum); ????return 0; } ```