Educational Codeforces Round 72 (Rated for Div. 2) C题

The Number Of Good Substrings

You are given a binary string s (recall that a string is binary if each character is either 0 or 1).

Let f(t) be the decimal representation of integer t written in binary form (possibly with leading zeroes). For example f(011)=3,f(00101)=5,f(00001)=1,f(10)=2,f(000)=0 and f(000100)=4.

The substring sl,sl+1,…,sr is good if r−l+1=f(sl…sr).

For example string s=1011 has 5 good substrings: s1…s1=1, s3…s3=1, s4…s4=1, s1…s2=10 and s2…s4=011.

Your task is to calculate the number of good substrings of string s.

You have to answer t independent queries.

Input

The first line contains one integer t (1≤t≤1000) — the number of queries.

The only line of each query contains string s (1≤|s|≤2⋅105), consisting of only digits 0 and 1.

It is guaranteed that ∑i=1t|si|≤2⋅105.

Output

For each query print one integer — the number of good substrings of string s.

---

二进制的一些理解和运用；

首先，要知道除了10（2）和1（1）可以不用前缀0以外，其他的数都需要前缀0，只是多少问题；

先记录每个1的位置和1前面的0的个数，然后枚举1后面20位（2^20已经比|s|大很多），算出每个位置与 1 组成的数所需要的前缀0，然后跟这个位置1的前缀0比较大小就行；

代码：

#include<bits/stdc++.h>
#define ll long long
#define pa pair<int,int>
#define lson k<<1
#define rson k<<1|1
//ios::sync_with_stdio(false);
using namespace std;
const int N=200100;
const int M=200100;
const ll mod=1e9+7;
struct Node{
	int post,date;
}a[N];
int solve(string p){
	int b=0;
	int ans=0;
	for(int i=p.length()-1;i>=0;i--){
		ans+=(p[i]-'0')*(1<<b);
		b++;
	}
	return ans;
}
int main(){
    ios::sync_with_stdio(false);
	int t;
	cin>>t;
	while(t--){
		ll ans=0;
		string s;
		cin>>s;
		int len=0,k=0;
		for(int i=0;i<s.length();i++){
			if(s[i]=='0') len++;
			else if(s[i]=='1'&&len){
				a[++k].date=len;
				a[k].post=i;
				len=0;
			}
		}
		for(int i=0;i<s.length();i++){
			if(s[i]=='1') ans++;
			if(i!=s.length()-1&&s[i]=='1'&&s[i+1]=='0') ans++;
		}
//		cout<<ans<<endl;
		for(int i=1;i<=k;i++){
			int l=a[i].post;
			string b="";
			for(int j=0;j<20;j++){
				if(l+j>=s.length()) break;
				b+=s[l+j];
				int c=solve(b)-j-1;
//				cout<<b<<" "<<c<<endl;
				if(c<=a[i].date&&c) ans++;
			}
		}
		cout<<ans<<endl;
	}
    return 0;
}