2189-Binary Number ZCMU

最新推荐文章于 2020-03-06 13:44:03 发布

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最新推荐文章于 2020-03-06 13:44:03 发布

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分类专栏：练习文章标签：贪心

本文链接：https://blog.youkuaiyun.com/ZCMU_2024/article/details/81273615

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本文介绍了一种算法，该算法将一个正整数从二进制形式转换为一系列的操作计数，目的是通过特定的操作步骤将任意正整数转换为1。具体操作包括：如果当前数为奇数，则加1；若为偶数，则除以2。文章提供了一个实现这一过程的C++代码示例。

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Description

Little walrus Fangy loves math very much. That's why when he is bored he plays with a number performing some operations.

Fangy takes some positive integer x and wants to get a number one from it. While x is not equal to 1, Fangy repeats the following action: if x is odd, then he adds 1 to it, otherwise he divides x by 2. Fangy knows that for any positive integer number the process ends in finite time.

How many actions should Fangy perform to get a number one from number x?

Input

The first line contains a positive integer x in a binary system. It is guaranteed that the first digit of x is different from a zero and the number of its digits does not exceed 106.

Output

Print the required number of actions.

Examples

Input

Output

Input

Output

Input

Output

Note

Let's consider the third sample. Number 101110 is even, which means that we should divide it by 2. After the dividing Fangy gets an odd number 10111 and adds one to it. Number 11000 can be divided by 2 three times in a row and get number 11. All that's left is to increase the number by one (we get 100), and then divide it by 2 two times in a row. As a result, we get 1.

解析

贪心

代码

#include<bits/stdc++.h>
using namespace std;
int main()
{
    char n[1005],len;
    int action,i,j;
    while(~scanf("%s",&n))
    {
        len=strlen(n)-1;
        action=0;
        while(len)
        {
           if(n[len]=='0')
           {
               action++;
               while(n[--len]=='0')
               {
                   action++;
               }
           }
           else
           {
               action+=2;
               while(--len&&n[len]=='1')
               {
                   action++;
               }
               if(len)
                 n[len]='1';
               else
                 action++;
           }
        }
        printf("%d\n",action);
    }
    return 0;
}