# 每次成功搜索后通過將元素加倍來重復搜索 > 原文： [https://www.geeksforgeeks.org/repeatedly-search-element-doubling-every-successful-search/](https://www.geeksforgeeks.org/repeatedly-search-element-doubling-every-successful-search/) 給定一個數組“ a []”和整數“ b”。 查找 b 是否存在于 a []中。 如果存在，則將 b 的值加倍并再次搜索。 我們重復這些步驟，直到找不到 b。 最后，我們返回 b 的值。 例子： ``` Input : a[] = {1, 2, 3} b = 1 Output :4 Initially we start with b = 1\. Since it is present in array, it becomes 2. Now 2 is also present in array b becomes 4 . Since 4 is not present, we return 4. Input : a[] = {1 3 5 2 12} b = 3 Output :6 ``` 問題來源：在 Yatra.com 在線測試中提出 1）對輸入數組進行排序。 2）繼續進行二分搜索并加倍，直到不存在該元素。 以下代碼在 STL 中使用 [binary_search（）](https://www.geeksforgeeks.org/binary-search-algorithms-the-c-standard-template-library-stl/) ## C++ ```cpp // C++ program to repeatedly search an element by // doubling it after every successful search #include <bits/stdc++.h> using namespace std; int findElement(int a[], int n, int b) { ????// Sort the given array so that binary search ????// can be applied on it ????sort(a, a + n); ????int max = a[n - 1]; // Maximum array element ????while (b < max) { ????????// search for the element b present or ????????// not in array ????????if (binary_search(a, a + n, b)) ????????????b *= 2; ????????else ????????????return b; ????} ????return b; } // Driver code int main() { ????int a[] = { 1, 2, 3 }; ????int n = sizeof(a) / sizeof(a[0]); ????int b = 1; ????cout << findElement(a, n, b); ????return 0; } ```