C++ 中陣列中所有成對連續元素的乘積
給定一個包含 n 個整數的陣列 arr[n],任務是找出所有成對連續元素的乘積。
陣列 arr[] 中的連續元素是,如果我們處於第 i 個元素,即 arr[i],則其連續元素將是 arr[i+1] 或 arr[i-1],因此乘積將是 arr[i] * arr[i+1] 或 arr[i] * arr[i-1]。
輸入
arr[] = {1, 2, 3, 4}
輸出
2, 6, 12
解釋
Splitting into pairs {1,2}, {2, 3}, {3, 4} Their results will be 1*2 = 2, 2*3 = 6, 3*4 = 12
輸入
arr[] = {9, 5, 1, 2, 6, 10}
輸出
45, 5, 2, 12, 60
解釋
Splitting into pairs {9, 5}, {5, 1}, {1, 2}, {2, 6}, {6, 10} Their results will be 9*5 = 45, 5*1 = 5, 1*2 = 2, 2*6=12, 6*10=60
用於解決此問題的演算法如下 −
從陣列的第 0 個元素開始迴圈,直到它小於 n-1。
對於每一個 i 檢查它的 i+1,逐一對 i 和 i+1 求積並列印結果。
演算法
Start Step 1→ Declare function to calculate product of consecutive elements void product(int arr[], int size) Declare int product = 1 Loop For int i = 0 and i < size – 1 and i++ Set product = arr[i] * arr[i + 1] Print product End Step 2 → In main() Declare int arr[] = {2, 4, 6, 8, 10, 12, 14 } Declare int size = sizeof(arr) / sizeof(arr[0]) Call product(arr, size) Stop
示例
#include <iostream> using namespace std; //functio to find the product of consecutive pairs void product(int arr[], int size){ int product = 1; for (int i = 0; i < size - 1; i++){ product = arr[i] * arr[i + 1]; printf("%d ", product); } } int main(){ int arr[] = {2, 4, 6, 8, 10, 12, 14 }; int size = sizeof(arr) / sizeof(arr[0]); printf("product is : "); product(arr, size); return 0; }
輸出
如果執行上述程式碼,它將生成以下輸出 −
product is : 8 24 48 80 120 168
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