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686 B

#include <bits/stdc++.h>
using namespace std;
const int N = 1000010;
int n, m, ne[N];
char s[N], p[N];
int main() {
cin >> n >> (p + 1) >> m >> (s + 1);
// 求模式串ne数组
for (int i = 2, j = 0; i <= n; i++) {
while (j && p[i] != p[j + 1]) j = ne[j];
if (p[i] == p[j + 1]) j++;
ne[i] = j;
}
// 求源串中模式串出现的每个位置
for (int i = 1, j = 0; i <= m; i++) {
while (j && s[i] != p[j + 1]) j = ne[j];
if (s[i] == p[j + 1]) j++;
if (j == n) {
printf("%d ", i - n);
j = ne[j]; // 继续搜索,重置 j=ne[j]
}
}
return 0;
}