这道题我们的做法是开两个vector，分别把a和b字符的下标存进去，然后遍历a字符，我们要求长度必须大于等于k，我们可以画个图，也就是说b的下标减a的下标必须大于等于k-1 也就是b的下标必须大于等于a的下标+k-1  我们用二分找出b数组里面大于等于这个值最小的数的下标，然后用nb减去这个下标就是满足要求的个数了

#include <iostream>
#include <vector>
using namespace std;
typedef long long LL;
string s;
LL ret;
vector<LL> a, b;
char ca, cb;
LL k;
int main()
{cin >> k >> s >> ca >> cb;LL ns = s.size();for (int i = 0; i < ns; i++){if (s[i] == ca) a.push_back(i);else if (s[i] == cb) b.push_back(i);}LL na = a.size();LL nb = b.size();for (int i = 0; i < na; i++){LL val = a[i] + k - 1;LL left = 0, right = nb - 1;while (left < right){LL mid = (left + right) / 2;if (b[mid] >= val) right = mid;else left = mid + 1;}if (b[left] >= val)ret +=  nb- left;}cout << ret << endl;
}