zcmu Problem E: Binary Number（模拟）

最新推荐文章于 2018-07-30 19:14:51 发布

芋圆西米露

最新推荐文章于 2018-07-30 19:14:51 发布

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本文链接：https://blog.youkuaiyun.com/qq_41117236/article/details/81271332

这是一道关于二进制数操作的问题。给定一个正整数x，小海象Fangy需要通过特定操作将其变为1。若x为奇数，加1；若x为偶数，除以2。任务是找出完成这一过程所需的步数。输入包含一个不超过10^6位且首位不为0的二进制数x，输出需要进行的操作次数。题解建议使用模拟方法解决。

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【题目】

Problem E: Binary Number

Time Limit: 2 Sec Memory Limit: 256 MB
Submit: 29 Solved: 13
[Submit][Status][Web Board]

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.

【题解】

题意：给你一串二进制，如果是奇数加1，如果是偶数除以2，问要做多少次操作才能得到1.

思路：模拟

【代码】

#include<bits/stdc++.h> 
using namespace std; 
char a[1000005]; 
int ans; 
void lowbit(int l,int r) 
{ 
    int f=1; 
    while(r>l) 
    { 
        if(a[r]=='1') 
        { 
            a[r]='0'; 
            r--; 
        } 
        else
        { 
            a[r]='1'; 
            f=0; 
            break; 
        } 
    } 
    if(f&&a[l]=='1') 
    { 
        ans++; 
    } 
} 
int main() 
{ 
    while(~scanf("%s",a)) 
    { 
        int l=strlen(a); 
        int i=0,j=l-1; 
        ans=0; 
        while(i<j) 
        { 
            ans++; 
            if(a[j]=='0') 
                j--; 
            else
                lowbit(i,j); 
        } 
        printf("%d\n",ans); 
    } 
    return 0; 
}