[Atcoder Code Festival 2017 Qual A Problem F] Squeezing Slimes

原题链接：https://code-festival-2017-quala.contest.atcoder.jp/tasks/code_festival_2017_quala_f

Squeezing Slimes

Problem Statement

There are A slimes lining up in a row. Initially, the sizes of the slimes are all 1.

Snuke can repeatedly perform the following operation.

Choose a positive even number M. Then, select M consecutive slimes and form M⁄2 pairs from those slimes as follows: pair the 1-st and 2-nd of them from the left, the 3-rd and 4-th of them, …, the (M−1)-th and M-th of them. Combine each pair of slimes into one larger slime. Here, the size of a combined slime is the sum of the individual slimes before combination. The order of the M⁄2 combined slimes remain the same as the M⁄2 pairs of slimes before combination.

Snuke wants to get to the situation where there are exactly N slimes, and the size of the i-th (1≤i≤N) slime from the left is ai. Find the minimum number of operations required to achieve his goal.

Note that A is not directly given as input. Assume A=a1+a2+…+aN.

Constraints

1≤N≤105
ai is an integer.
1≤ai≤ $10^9$

Input

Input is given from Standard Input in the following format:

N
a1 a2 … aN

Output

Print the minimum number of operations required to achieve Snuke’s goal.

Sample Input 1

2
3 3

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Sample Output 1

One way to achieve Snuke’s goal is as follows. Here, the selected slimes are marked in bold.

$(1, 1, 1, 1, 1, 1) → (1, 2, 2, 1)$
$(1, 2, 2, 1) → (3, 3)$

Sample Input 2

4
2 1 2 2

Sample Output 2

One way to achieve Snuke’s goal is as follows.

$(1, 1, 1, 1, 1, 1, 1) → (2, 1, 1, 1, 1, 1)$
$(2, 1, 1, 1, 1, 1) → (2, 1, 2, 2)$

Sample Input 3

1
1

Sample Output 3

Sample Input 4

10
3 1 4 1 5 9 2 6 5 3

Sample Output 4

题目大意

有 $n$ 个 $1$ 组成的数列，每次可以选择连续偶数个数字。将第 $1$ 个和第 $2$ 个合并（求和），第 $3$ 个和第 $4$ 个合并，以此类推。要求最终结果是输入的数列。

至少要多少次操作能达到目标序列。

题解

毕克讲的全都是短代码神题 $orz$ 。

我们知道结束状态的数列，所以考虑倒推，每次选一些数分成两份，那么对于一个数 $x$ ，最快的拆分就是按二进制拆分，我们需要找到最大的 $mi$ 使得 $2^{mi}\le x$ 。

如果 $2^{mi}=x$ ，就需要 $mi$ 次操作将 $x$ 拆分；如果 $2^{mi}<x$ ，就需要再来一次。

考虑 $x$ 的前一个元素的拆分次数为 $now$ ，当 $now=mi$ 时，直接将上一个元素的操作区间扩大将 $x$ 覆盖就行了；如果 $now<mi$ ，再扩大上一个元素的操作区间时，我们还需要更多的操作划分 $x$ ； $now<mi$ 时不再新增操作。

代码

#include<bits/stdc++.h>
using namespace std;
int n,x,mi,ans,dlt,now;
void in(){scanf("%d",&n);}
void ac()
{
    for(int i=1;i<=n;++i)
    {
        dlt=0;
        scanf("%d",&x);
        mi=(int)log2(x);
        if(x!=(1<<mi))dlt=1;
        if(now<mi)ans+=mi-now,now=mi;
        if(now>mi)now=mi,dlt=0;
        if(dlt)++ans,++now;
    }
    printf("%d",ans);
}
int main(){in();ac();}