cf165c Another Problem on Strings

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最新推荐文章于 2024-02-22 14:18:53 发布

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本文介绍了一种算法，用于解决在一个二进制字符串中找到包含恰好k个字符'1'的所有不同子串的问题。该算法通过遍历输入字符串并使用额外数组跟踪'0'字符之间的距离来高效地计算结果。

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A string is binary, if it consists only of characters "0" and "1".

String v is a substring of string w if it has a non-zero length and can be read starting from some position in string w. For example, string "010" has six substrings: "0", "1", "0", "01", "10", "010". Two substrings are considered different if their positions of occurrence are different. So, if some string occurs multiple times, we should consider it the number of times it occurs.

You are given a binary string s. Your task is to find the number of its substrings, containing exactly k characters "1".

Input

The first line contains the single integer k (0 ≤ k ≤ 106). The second line contains a non-empty binary string s. The length of s does not exceed 106 characters.

Output

Print the single number — the number of substrings of the given string, containing exactly k characters "1".

Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64dspecifier.

Sample test(s)

input
1
1010

output
6

input
2
01010

output
4

input
100
01010

output
0

Note

In the first sample the sought substrings are: "1", "1", "10", "01", "10", "010".

In the second sample the sought substrings are: "101", "0101", "1010", "01010".

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;

long long a[1111111];
char s[1111111];

int main()
{
    long long k;
    memset(a,0,sizeof(a));
    memset(s,0,sizeof(s));
    cin>>k;
    scanf("%s",s);
    long long vis=0,res=0,ans=0,i;
    long long len=strlen(s);
    for(i=0;i<len;i++)
    {
        if(s[i]=='1')
        {
            a[res++]=vis;
            vis=0;
        }
        else
            vis++;
    }
    a[res]=vis;
    //for(i=0;i<=res;i++)  cout<<a[i]<<endl;
    if(k==0)
    {
        for(i=0;i<=res;i++)
        {
            while(a[i]>=1)
            {
                ans+=a[i];
                a[i]--;
            }
        }
    }
    else
    for(i=0;i+k<=res;i++)
        ans+=(a[i]+1)*(a[i+k]+1);
    cout<<ans<<endl;


    return 0;
}