题解 | nico和niconiconi

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;

typedef long long ll;

// 检查从位置start开始的子串是否等于pattern
bool check(const string& s, int start, const string& pattern) {
    for (size_t i = 0; i < pattern.size(); i++) {
        if (s[start + i] != pattern[i]) return false;
    }
    return true;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int n;
    ll a, b, c;
    string s;
    cin >> n >> a >> b >> c >> s;
    
    vector<ll> dp(n + 1, 0);
    
    for (int i = 0; i <= n; i++) {
        if (i > 0) {
            dp[i] = max(dp[i], dp[i - 1]);
        }
        
        // 向前看，检查从当前位置开始的匹配
        if (i + 4 <= n && check(s, i, "nico")) {
            dp[i + 4] = max(dp[i + 4], dp[i] + a);
        }
        if (i + 6 <= n && check(s, i, "niconi")) {
            dp[i + 6] = max(dp[i + 6], dp[i] + b);
        }
        if (i + 10 <= n && check(s, i, "niconiconi")) {
            dp[i + 10] = max(dp[i + 10], dp[i] + c);
        }
    }
    
    cout << dp[n] << "\n";
    
    return 0;
}