[topcoder]IncrementAndDoubling

http://community.topcoder.com/stat?c=problem_statement&pm=12790&rd=15708

这道题只有两个操作，一是加一，二是数组所有元素乘以二，求最少操作次数从全0数组变成目标数组。我的方法是从目标数组反推全0数组，如果有奇数就变成偶数，全偶数就除以二。这个方法是可以过的，就是效率拙计：

import java.util.*;

public class IncrementAndDoubling

{

    public int getMin(int[] A)

    {

        int count = 0;

        int len = A.length;

        while (true)

        {

            boolean done = true;

            for (int i = 0; i < len; i++)

            {

                if (A[i] % 2 == 1)

                {

                    A[i]--;

                    count++;

                }

                if (A[i] != 0)

                    done = false;

            }

            if (done) break;

            for (int i = 0; i < len; i++)

            {

                A[i] = A[i] / 2;

            }

            count++;

        }

        return count;

    }

}

但更好的方法如标程里所介绍，观察到除以2就是移位，0和1的转换。那么最后的结果其实是数组里所有二进制1的个数，加上最长的二进制表示长度。

int getMin(vector<int> desiredArray)

{

    int mx = 1;  // '0' has length 1

    int sum = 0;

    for (int x: desiredArray) {

        int c = 0;

        // extract bits from x:

        while (x > 0) {

            c++;

            sum += x % 2; //the last bit

            x /= 2;

        }

        mx = std::max(mx, c);

        // the number c of bits is wrong for '0', but it doesn't matter

        // because mx is initially set to 1.

    }

    return mx - 1 + sum;

}