Tokio Marine & Nichido Fire Insurance Programming Contest 2023

打一半说不计分了，刚好天梯赛打完就不想写了，后面不先把写过的写个题解把

Problem Statement

A

You are given a string S of length N consisting of three kinds of characters: ., |, and *. S contains exactly two | and exactly one *.

Determine whether the * is between the two |, and if so, print in; otherwise, print out.

More formally, determine whether one of the characters before the * is | and one of the characters after the * is |.

Constraints

• 3≤N≤100
• N is an integer.
• S is a string of length N consisting of ., |, and *.
• S contains exactly two |.
• S contains exactly one *.

---

Input

The input is given from Standard Input in the following format:

N
S

Output

Print a single line containing in if the * is between the two |, and out otherwise.

代码

void solve()
{
    cin >> n;
    string s;
    cin >> s;
    int x = 0;
    for(int i = 0;i < n;i ++)
    {
        if(s[i] == '*')
        {
            x = i;
        }
    }
        for(int i = x-1;i >= 0;i --)
        {
            if(s[i] == '|')
            f = false;
        }
        for(int i = x+1;i < n;i ++)
        {
            if(s[i] == '|')
            f1 = false;
        }
    if(!f&&!f1)
    cout << "in" << '\n';
    else
    cout << "out" << '\n';
}

Problem Statement

B

N players with ID numbers 1,2,…,N are playing a card game.
Each player plays one card.

Each card has two parameters: color and rank, both of which are represented by positive integers.
For i=1,2,…,N, the card played by player i has a color Ci​ and a rank Ri​. All of R1​,R2​,…,RN​ are different.

Among the N players, one winner is decided as follows.

• If one or more cards with the color T are played, the player who has played the card with the greatest rank among those cards is the winner.
• If no card with the color T is played, the player who has played the card with the greatest rank among the cards with the color of the card played by player 11 is the winner. (Note that player 1 may win.)

Print the ID number of the winner.

Constraints

• 2≤N≤2×105
• 1≤T≤109
• 1≤Ci​≤109
• 1≤Ri​≤109
• i=j⟹Ri​=Rj​
• All values in the input are integers.

void solve()
{
    cin >> n >> m;
    map<int,int>mp;
    for(int i = 1;i <= n;i ++)
    {
        cin >> a[i];
    }
    for(int i = 1;i <= n;i ++)
    {
        cin >> b[i];
    }
    for(int i = 1;i <= n;i ++)
    {
        if(!mp[a[i]])
        mp[a[i]] = i;
        else
        {
            if(b[mp[a[i]]] < b[i])
            mp[a[i]] = i;
        }
     }
    if(mp[m])
    cout << mp[m] << '\n';
    else
    cout << mp[a[1]] << '\n';
    }
    signed main()
    {
        int tt = 1;
    //sc(tt);
    while(tt--)
    {
        solve();
    }
}

C

For a positive integer L, a level-L dango string is a string that satisfies the following conditions.

• It is a string of length L+1 consisting of o and -.
• Exactly one of the first character and the last character is -, and the other L characters are o.

For instance, ooo- is a level-3 dango string, but none of -ooo-, oo, and o-oo- is a dango string (more precisely, none of them is a level-L dango string for any positive integer L).

You are given a string S of length N consisting of the two characters o and -. Find the greatest positive integer X that satisfies the following condition.

• There is a contiguous substring of S that is a level-X dango string.

If there is no such integer, print -1.

Constraints

• 1≤N≤2×105
• S is a string of length N consisting of o and -.

代码

void solve()
{
    cin >> n;
    string s;
    cin >> s;
    int ans = 0,num = 0;
    for(int i = 1;i < n;i ++)
    {
        if(s[i] == 'o'&&s[i-1] == '-')
        {
            for(int j = i;j < n;j ++)
            {
                if(s[j] == 'o')
                num++;
                else
                break;
            }
        if(i + num - 1< n)
        {
            ans = max(ans,num);
        }
        num = 0;
        }
   }
    for(int i = 0;i < n;i ++)
    {
        if(s[i] == 'o')
        {
            num++;
        }
        else
        {
            ans = max(ans,num);
            num = 0;
        }
    }
    if(ans)
        cout << ans << '\n';
    else
        cout << -1 << '\n';
    }
    signed main()
    {
    int tt = 1;
    //sc(tt);
    while(tt--)
    {
        solve();
    }
}