A. BowWow and the Timetable
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
In the city of Saint Petersburg, a day lasts for 2100 minutes. From the main station of Saint Petersburg, a train departs after 1 minute, 4 minutes, 16 minutes, and so on; in other words, the train departs at time 4k for each integer k≥0. Team BowWow has arrived at the station at the time s and it is trying to count how many trains have they missed; in other words, the number of trains that have departed strictly before time s. For example if s=20, then they missed trains which have departed at 1, 4 and 16. As you are the only one who knows the time, help them!
Note that the number s will be given you in a binary representation without leading zeroes.
Input
The first line contains a single binary number s (0≤s<2100) without leading zeroes.
Output
Output a single number — the number of trains which have departed strictly before the time s.
Examples
inputCopy
100000000
outputCopy
4
inputCopy
101
outputCopy
2
inputCopy
10100
outputCopy
3
Note
In the first example 1000000002=25610, missed trains have departed at 1, 4, 16 and 64.
In the second example 1012=510, trains have departed at 1 and 4.
The third example is explained in the statements.
题意：
给你一个二进制的数x，问有多少个4^i 小于x
思路：
因为给的是二进制数，
我们知道一个二进制的数右移一位是除以2，那么右移两位就是除以4，
那么我们只需要看二进制数能右移2位几次即可，这个只需要二进制的长度即可判定。
又因为要求严格的小于x，那么如果是1000000，这样的数字，其实有效到的是 111111 ,即有效位数是减去1的。
那么直接有效长度/2就是答案
细节见代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int main()
{
//freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
//freopen("D:\\common_text\\code_stream\\out.txt","w",stdout);
string str;
cin>>str;
ll ans=0ll;
int flag=1;
for(int i=1;i<sz(str);++i)
{
if(str[i]!='0')
{
flag=0;
}
}
int len=sz(str);
if(flag)
{
len--;
}
ans=(len+1)/2;
cout<<ans<<endl;
return 0;
}
inline void getInt(int* p) {
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '\n');
if (ch == '-') {
*p = -(getchar() - '0');
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 - ch + '0';
}
}
else {
*p = ch - '0';
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 + ch - '0';
}
}
}