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 "\n"
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;
}