退役啦，接下来的博客全是图一乐啦 

E - Joint Two Strings

题意

思路

统计两个指针的方案数一定是枚举一个，统计另一个

然后因为拼起来之后要包含 t 这个字符串，隐隐约约会感觉到和前缀后缀子序列有关

考虑预处理每个 s[i] 的最长公共前缀子序列 和 最长公共后缀子序列，接下来问题就变成：

满足ai + bj >= m有多少对 (i, j)，这个直接值域统计一下即可

#include <bits/stdc++.h>#define int long longconstexpr int N = 5e5 + 10;
constexpr int M = 1e4 + 10;
constexpr int mod = 1e9 + 7;
constexpr int Inf = 0x3f3f3f3f;std::string t, s[N];int n;
int pre[N], suf[N];
int mp[N];void solve() {std::cin >> n >> t;int m = t.size();t = " " + t;for (int i = 1; i <= n; i ++) {std::cin >> s[i];s[i] = " " + s[i];}for (int i = 1; i <= n; i ++) {int len = s[i].size() - 1;pre[i] = 1;for (int j = 1; j <= len; j ++) {if (pre[i] <= m && t[pre[i]] == s[i][j]) {pre[i] += 1;}}pre[i] -= 1;suf[i] = m;for (int j = len; j >= 1; j --) {if (suf[i] >= 1 && t[suf[i]] == s[i][j]) {suf[i] -= 1;}}suf[i] += 1;mp[suf[i]] += 1;}for (int i = 2; i <= m + 1; i ++) {mp[i] += mp[i - 1];}int ans = 0;for (int i = 1; i <= n; i ++) {ans += mp[pre[i] + 1];}std::cout << ans << "\n";
}
signed main(){std::ios::sync_with_stdio(false);std::cin.tie(nullptr);int t = 1;while(t --) {solve();}return 0;
}