TopCoder SRM 596 DIV 1 250

Problem Statement
	You have an array with N elements. Initially, each element is 0. You can perform the following operations: Increment operation: Choose one element of the array and increment the value by one. Doubling operation: Double the value of each element. You are given a vector <int> desiredArray containing N elements. Compute and return the smallest possible number of operations needed to change the array from all zeros to desiredArray.
Definition
	Class: IncrementAndDoubling Method: getMin Parameters: vector <int> Returns: int Method signature: int getMin(vector <int> desiredArray) (be sure your method is public)
Limits
	Time limit (s): 2.000 Memory limit (MB): 64
Constraints
-	desiredArray will contain between 1 and 50 elements, inclusive.
-	Each element of desiredArray will be between 0 and 1,000, inclusive.
Examples
0)
	{2, 1} Returns: 3 One of the optimal solutions is to apply increment operations to element 0 twice and then to element 1 once. Total number of operations is 3.
1)
	{16, 16, 16} Returns: 7 The optimum solution looks as follows. First, apply an increment operation to each element. Then apply the doubling operation four times. Total number of operations is 3+4=7.
2)
	{100} Returns: 9
3)
	{0, 0, 1, 0, 1} Returns: 2 Some elements in desiredArray may be zeros.
4)
	{123, 234, 345, 456, 567, 789} Returns: 40
5)
	{7,5,8,1,8,6,6,5,3,5,5,2,8,9,9,4,6,9,4,4,1,9,9,2,8,4,7,4,8,8,6,3,9,4,3,4,5,1,9,8,3,8,3,7,9,3,8,4,4,7} Returns: 84

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集训队题解

给出一个序列，n个元素，初始为0。操作有二：

1.给某个数加1

2.给所有数乘2

求变成目标序列的最小次数。

 #include <bits/stdc++.h>
 using namespace std;

 class IncrementAndDoubling {
 public:
     int getMin(vector<int> desiredArray) {
         int cnt = , maxi = ;
         for (auto i : desiredArray)
             for (int j = ; i >> j; ++j)
                 maxi = max(maxi, j), cnt += (i >> j) & ;
         return cnt + maxi;
     }
 };

@Author: YouSiki