CodeForces

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”.

这玩意就是找几个零几个一非常容易。。
有点坑的是运算过程一定要*1LL不然过程中就爆了int

#include<iostream>#include<string>#include<algorithm>using namespace std;string q;int n,m=0;int yi[1000001];int main(){    cin >> n >> q;//  n = 0;//  q = "0";//  for (int a = 1; a <= 99999; a++)q += "0";    for (int a = 0; a < q.size(); a++)if (q[a] == '1')yi[++m] = a+1;    if (n == 0)    {        long long we = 0;        for (int a = 1; a <= m; a++)we += 1LL*((yi[a] - yi[a - 1] - 1) + 1)*(yi[a] - yi[a - 1] - 1) / 2;        we += 1LL * (q.size() - yi[m])*(q.size() - yi[m] + 1) / 2;        cout << we;        return 0;    }    long long jg = 0;    for (int a = 1, b = n; a <= m && b <= m; a++, b++)    {        if (b == m)        {            jg += 1LL * (yi[a] - yi[a - 1])*(q.size() - yi[b]+1);            continue;        }        jg+= 1LL*(yi[a] - yi[a - 1])*(yi[b+1] - yi[b]);    }    cout << jg;}