codeforces 666C Codeword

codeforces 666C Codeword

题意

q个询问，一种询问是给你一个字符串，还有一种是问长度为n的，包含当前字符串为子序列的字符串有多少个。

题解

容易写出式子，但是不好化简。
观察一下可以知道q个询问的字符串长度也就根号种。

代码

#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define rep(i, a, b) for(int i=(a); i<(b); i++)
#define sz(a) (int)a.size()
#define de(a) cout << #a << " = " << a << endl
#define dd(a) cout << #a << " = " << a << " "
#define all(a) a.begin(), a.end()
#define endl "
"
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
//---

const int N = 101010, P = 1e9+7;

int pw[N], jc[N], in[N], ans[N];
vector<pii> Q[N];

inline int mul(int a, int b) {
	return a * 1ll * b % P;
}
inline int add(int a, int b) {
	int res = a+b;
	if(res >= P) res -= P;
	return res;
}

inline int kpow(int a, int b) {
	int res = 1;
	while(b) {
		if(b&1) res = mul(res, a);
		a = mul(a, a);
		b >>= 1;
	}
	return res;
}

inline void init() {
	pw[0] = 1;
	rep(i, 1, N) pw[i] = mul(pw[i-1], 25);
	jc[0] = 1;
	rep(i, 1, N) jc[i] = mul(jc[i-1], i);
	in[N-1] = kpow(jc[N-1], P-2);
	for(int i = N-2; ~i; --i) in[i] = mul(in[i+1], i+1);
}

inline int C(int n, int m) {
	if(n < m) return 0;
	return mul(jc[n], mul(in[m], in[n-m]));
}

int main() {
	std::ios::sync_with_stdio(false);
	std::cin.tie(0);
	init();
	int n, m, q;
	string s;
	cin >> q >> s;
	m = sz(s);
	rep(i, 0, q) {
		int x;
		cin >> x;
		if(x == 1) {
			cin >> s;
			m = sz(s);
		} else {
			cin >> n;
			Q[m].pb(mp(n, i));
		}
	}
	rep(m, 0, N) if(sz(Q[m])) {
		sort(all(Q[m]));
		int i = m, res = 1;
		for(auto t : Q[m]) {
			int n = t.fi;
			while(i < n) {
				++i;
				res = mul(res, 26);
				res = add(res, mul(C(i-1, m-1), pw[i - m]));
			}
			ans[t.se] = n >= m ? res : 0;
			++ans[t.se];
		}
	}
	rep(i, 0, q) if(ans[i]) {
		cout << ans[i]-1 << endl;
	}
	return 0;
}