Submission #17919427

---

Source Code Expand

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
constexpr int Inf = 1000000030;
constexpr ll INF= 2000000000000000000;
constexpr ll MOD = 998244353;
const double PI = 3.1415926535897;
typedef pair<int,int> P;
typedef pair<int,P> PP;

template<class T> inline bool chmax(T &a, T b) {
    if (a < b) {
        a = b;
        return 1;
    }
    return 0;
}

template<class T> inline bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return 1;
    }
    return 0;
}

ll mod(ll val, ll M) {
    val = val % M;
    if(val < 0) {
        val += M;
    }
    return val;
}

template<typename T>
T RS(T N, T P, T M){
    if(P == 0) {
        return 1;
    }
    if(P < 0) {
        return 0;
    }
    if(P % 2 == 0){
        ll t = RS(N, P/2, M);
        if(M == -1) return t * t;
        return t * t % M;
    }
    if(M == -1) {
        return N * RS(N,P - 1,M);
    }
    return N * RS(N, P-1, M) % M;
}

bool ng[27][27];
int dp[300010][27]; //dp[i][j] = i文字目までで文字jが最後に来る場合の数（ただしj = 26は空文字列）

int main() {
    int L;
    cin >> L;
    string S;
    cin >> S;
    int M;
    cin >> M;
    for(int i = 0;i < 27;i++) {
        for(int j = 0;j < 27;j++) {
            ng[i][j] = false;
        }
    }
    for(int i = 0;i < M;i++) {
        char A,B;
        cin >> A >> B;
        ng[A - 'A'][B - 'A'] = true;
    }
    for(int i = 0;i < 300010;i++) {
        for(int j = 0;j < 27;j++) {
            dp[i][j] = 0;
        }
    }
    dp[0][26] = 1;
    for(int i = 0;i < L;i++) {
        for(int j = 0;j <= 26;j++) {
            if(!ng[j][S.at(i) - 'A']) {
                dp[i + 1][S.at(i) - 'A'] += dp[i][j];
                dp[i + 1][S.at(i) - 'A'] %= 10000000;
            }
            if(j != S.at(i) - 'A') {
                dp[i + 1][j] += dp[i][j];
            }
        }
    }
    int ret = 0;
    for(int i = 0;i < 26;i++) {
        ret += dp[L][i];
        ret %= 10000000;
    }
    cout << ret << endl;
}

Submission Info

Judge Result