# 計算總數小于給定值的三元組 > 原文： [https://www.geeksforgeeks.org/count-triplets-with-sum-smaller-that-a-given-value/](https://www.geeksforgeeks.org/count-triplets-with-sum-smaller-that-a-given-value/) 給定一個由不同整數組成的數組和一個求和值。 查找總和小于給定總和值的三元組的數量。 預期時間復雜度為 `O(n^2)`。 **示例**： ``` Input : arr[] = {-2, 0, 1, 3} sum = 2. Output : 2 Explanation : Below are triplets with sum less than 2 (-2, 0, 1) and (-2, 0, 3) Input : arr[] = {5, 1, 3, 4, 7} sum = 12. Output : 4 Explanation : Below are triplets with sum less than 12 (1, 3, 4), (1, 3, 5), (1, 3, 7) and (1, 4, 5) ``` 一個**簡單解決方案**是運行三個循環，一個接一個地考慮所有三元組。 對于每個三元組，如果三元組的總和小于給定的總和，則比較總和和增量計數。 ## C++ ```cpp // A Simple C++ program to count triplets with sum smaller // than a given value #include<bits/stdc++.h> using namespace std; int countTriplets(int arr[], int n, int sum) { ????// Initialize result ????int ans = 0; ????// Fix the first element as A[i] ????for (int i = 0; i < n-2; i++) ????{ ???????// Fix the second element as A[j] ???????for (int j = i+1; j < n-1; j++) ???????{ ???????????// Now look for the third number ???????????for (int k = j+1; k < n; k++) ???????????????if (arr[i] + arr[j] + arr[k] < sum) ???????????????????ans++; ???????} ????} ????return ans; } // Driver program int main() { ????int arr[] = {5, 1, 3, 4, 7}; ????int n = sizeof arr / sizeof arr[0]; ????int sum = 12; ????cout << countTriplets(arr, n, sum) << endl; ????return 0; } ```