C - Another Problem on Strings CodeForces - 165C

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最新推荐文章于 2024-08-02 17:51:14 发布

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本文介绍了一种算法，用于计算给定二进制字符串中包含恰好k个字符'1'的子串数量。该算法首先定位所有'1'的位置，并通过计算相邻'1'之间的'0'数量来确定子串的数目。

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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 %I64d specifier.

Example
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”.

解题思路：如果找n（非0）个1组成的子串，需要先找到n个1的子串，此时的子串左边a个0，右边b个0，这样符合条件的解为（a+1）（b+1）。然后循环寻找。如果找0个1的子串，先找到排列的m个0，这m个0符合条件的解为m（m-1）/2，然后循环找完全部。

#include<bits/stdc++.h>
using namespace std;
long long num,ans=0,sor[1000005];
char s[1000006];
int main(){
    cin>>num;
    cin>>s; 
    long long i = 0,k=1;
    while(s[i++]!='\0')
        if(s[i-1]-'0'==1) sor[k++] = i; 
    sor[0] = 0;
    sor[k] = strlen(s)+1;
    if(num){
        for(int i = num;i < k;i++)
            ans = ans + (sor[i-num+1]-sor[i-num])*(sor[i+1]-sor[i]);

    }
    else{
        for(long long i = 1;i <= k;i++){
            ans += (sor[i]-sor[i-1]-1)*(sor[i]-sor[i-1])/2; 
        }   
    }
    cout<<ans<<endl;
    return 0;
}