Codeforces Round #444 (Div. 2)A. Div. 64【进制思维】

A. Div. 64

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Top-model Izabella participates in the competition. She wants to impress judges and show her mathematical skills.

Her problem is following: for given string, consisting of only 0 and 1, tell if it's possible to remove some digits in such a way, that remaining number is a representation of some positive integer, divisible by 64, in the binary numerical system.

Input

In the only line given a non-empty binary string s with length up to 100.

Output

Print «yes» (without quotes) if it's possible to remove digits required way and «no» otherwise.

Examples

input

100010001

output

yes

input

100

output

no

Note

In the first test case, you can get string 1 000 000 after removing two ones which is a representation of number 64 in the binary numerical system.

You can read more about binary numeral system representation here: https://en.wikipedia.org/wiki/Binary_system

【题意】： 给你一个只包含0和1的字符串 ， 判断这个 字符串 能不能通过删除一些1或者0 使得剩下的数字是一个整数的二进制表示，并且能够被二进制的 64 整除。

【分析】：64 的二进制表示是：1000000（2） ， 从给出的数字的最高位向后找到第一个 1 然后向后统计 0 的个数 num， num大于等于 6 则输出 yes 否则输出 no

【代码】：

#include <bits/stdc++.h>

using namespace std;
char s[11000];
int main()
{
    scanf("%s",s);
    int flag=0,cnt=0;
    for(int i=0;i<strlen(s);i++)
    {
        if(flag==0 && s[i]=='1')
        {
            flag=1;
        }
        if(flag)
        {
            if(s[i]=='0')
            {
                cnt++;
            }
        }
    }
    if(cnt>=6)
        printf("yes
");
    else
        printf("no
");
    return 0;
}