题解 | 小红的01子序列构造（easy）

#include <bits/stdc++.h>

using namespace std;

int main() {
    long long n, k;
    cin >> n >> k;
    string s;
    cin >> s;
    s = '.' + s;
    vector<long long> qzh(n + 1, 0);
    for (int i = 1; i <= n; i++) {
        qzh[i] = qzh[i - 1] + s[i] - '0';
    }
    long long z01 = 0;
    long long jj0 = 0; //j前边0零的数量
    int i = 1, j = 2;
    if (s[i] == '0' && s[j] == '1')z01++;
    for (int k = i; k <= j; k++) {
        if (s[k] == '0')jj0++;
    }
    while (i < n && j<=n) {
        if (z01 == k) {
            break;
        }
        while (z01 > k && i < j) {
            if (s[i] == '0') {
                z01 -= qzh[j] - qzh[i];
                jj0--;
            }
            i++;
        }
        while (z01 < k) {
            j++;
            if(j>n)break;
            if (s[j] == '1') {
                z01 += jj0;
            } else if (s[j] == '0')jj0++;
        }
    }
    if (z01 == k) {
        cout << i << ' ' << j;
    } else cout << -1;
}