2025牛客寒假训练营3部分题解（vegetable dog）

1 前言：

前几天刚说今年的寒假训练营变简单了，这次的训练营3马上就给了本蒟蒻当头一棒，导致这次只做出来了6题（doge）

2 题解

A 智乃的博弈游戏

签到题。判断n的奇偶性即可。

void solve()
{
    int n;
    cin >> n;
    if (n % 2 == 0)
    cout << "No" << endl;
    else
    cout << "Yes" << endl;
}

M 智乃的牛题

签到题。用桶判断指定字符是否出现指定次数即可。

void solve()
{
    string s;
    cin >> s;
    vector<int>a(125,0);
    for  (int i = 0;i < 125;i++)
    {
        a[s[i]]++;
    }
    if (a[99] == 1 && a[100] == 1 && a[101] == 1 && a[110] == 1 && a[111] == 2 && a[114] == 1 && a[119] == 1)
    {
        cout << "happy new year" << endl;
    }
    else
    cout << "I AK IOI" << endl;
}

F 智乃的捉迷藏

签到题。本题数据很弱，如果找不到规律可以六重循环判断，也可以根据规律判断。

void solve()
{
    int n, a, b, c;
    cin >> n >> a >> b >> c;
    bool judge = false;
    for (int x1 = 0; x1 <= n; x1++) {
        for (int x2 = 0; x2 <= n - x1; x2++) {
            for (int x3 = 0; x3 <= n - x1 - x2; x3++) {
                for (int x4 = 0; x4 <= n - x1 - x2 - x3; x4++) {
                    for (int x5 = 0; x5 <= n - x1 - x2 - x3 - x4; x5++) {
                        int x6 = n - x1 - x2 - x3 - x4 - x5;
                        if (x6 >= 0) {
                            int sum1 = x1 + x2 + x3;
                            int sum2 = x3 + x4 + x5;
                            int sum3 = x5 + x6 + x1;
                            if (sum1 == a && sum2 == b && sum3 == c) {
                                judge = true;
                                break;
                            }
                        }
                    }
                    if (judge) break;
                }
                if (judge) break;
            }
            if (judge) break;
        }
        if (judge) break;
    }
    if (judge) {
        cout << "Yes" << endl;
    } 
    else {
        cout << "No" << endl;
    }
}

void solve1()
{
    int n,a,b,c;
    cin >> n >> a >> b >> c;
    int sum = a + b + c;
    if (sum >= n && sum <= 2 * n)
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
}

L 智乃的三角遍历

简单题。本题找到规律后手动打表即可。（本蒟蒻有点菜，赛时没有想出优雅的递归写法（哭哭））

void solve()
{
    int n;
    cin >> n;
    int a1[1] = {1};
    int a2[2] = {2,3};
    int a3[3] = {4,5,6};
    int a4[4] = {7,8,9,10};
    int a5[5] = {11,12,13,14,15};
    int a6[6] = {16,17,18,19,20,21};
    int a7[7] = {22,23,24,25,26,27,28};
    int a8[8] = {29,30,31,32,33,34,35,36};
    if (n == 1)
    cout << "1 2 3 1" << endl;
    if (n == 2)
    cout << "1 3 6 5 4 2 5 3 2 1" << endl;
    if (n == 3)
    cout << "1 3 6 10 9 8 7 4 8 5 9 6 5 4 2 5 3 2 1" << endl;
    if (n == 4)
    cout << " 1 3 6 10 15 14 13 12 11 7 12 8 13 9 14 10 9 8 7 4 8 5 9 6 5 4 2 5 3 2 1" << endl;
    if (n == 5)
    cout << "1 3 6 10 15 21 20 19 18 17 16 11 17 12 18 13 19 14 20 15 14 13 12 11 7 12 8 13 9 14 10 9 8 7 4 8 5 9 6 5 4 2 5 3 2 1" << endl;
    if (n == 6)
    cout << "1 3 6 10 15 21 28 27 26 25 24 23 22 16 23 17 24 18 25 19 26 20 27 21 20 19 18 17 16 11 17 12 18 13 19 14 20 15 14 13 12 11 7 12 8 13 9 14 10 9 8 7 4 8 5 9 6 5 4 2 5 3 2 1" << endl;
    if (n == 7)
    cout << "1 3 6 10 15 21 28 36 35 34 33 32 31 30 29 22 30 23 31 24 32 25 33 26 34 27 35 28 27 26 25 24 23 22 16 23 17 24 18 25 19 26 20 27 21 20 19 18 17 16 11 17 12 18 13 19 14 20 15 14 13 12 11 7 12 8 13 9 14 10 9 8 7 4 8 5 9 6 5 4 2 5 3 2 1" << endl;
}

C 智乃的Notepad(easy)

本题的hard版本为多组查询，本蒟蒻只能想到字典树，然后就想不下去了，所以在easy版本的单组查询时发现只需要将字符串按字典序排好，然后遍历一遍，将size累加即可。注意最后要将maxsize处理一下。

string s[100005];
void solve()
{
    int n, m, i, j, l, r;
    cin >> n >> m;
    for (i = 1; i <= n; i++)
    {
        cin >> s[i];
    }
    cin >> l >> r;
    sort(s + l, s + r + 1);

    int cnt = 0;
    int maxsize = 0;
    for (i = l; i <= r; i++)
    {
        maxsize = max(maxsize, (int)s[i].size());
        if (i == l)
        {
            cnt += s[i].size();
            continue;
        }
        for (j = 0; j < s[i].size() && j < s[i - 1].size(); j++)
        {
            if (s[i][j] != s[i - 1][j])
            {
                break;
            }
        }

        cnt += s[i].size() - j;
        cnt += s[i - 1].size() - j;
    }
    cnt += s[n].size();
    cnt -= maxsize;
    cout << cnt << endl;
}

E 智乃的小球

简单-中等题。本题由于所有小球的质量，速度都相同，且发生完全弹性碰撞，所以所有小球碰撞后只会改变运动方向，速度不变。所以我们可以认为两小球碰撞后互不影响。
根据此性质，我们可以将小球分为从左往右和从右往左两类，然后二分碰撞时间来判断在该时间是否发生第k次碰撞即可。
(其实本蒟蒻一开始没往二分去想，但是注意到碰撞时间只有整数或者0.5，但是题目却给了精度判断，所以推测出二分时间的解法)

void solve()
{
    int n, k;
    cin >> n >> k;
    for (int i = 1; i <= n; i++)
    {
        int p, v;
        cin >> p >> v;
        mp[p] = v;
    }
    double l = 0, r = 1e9 + 7;
    double mid;
    double e = 1e-7;
    while (r - l > e) 
    { 
        mid = (l + r) / 2;
        if (mid > 1e9)
        {
            printf("No\n");
            return;
        }
        if (midsearch(mid) >= k) 
            r = mid;  
        else 
            l = mid;  
    }
   mid = (l + r) / 2;  
   printf("Yes\n");
   printf("%.12lf\n", mid);
}

3 总结

我还是太菜了555，是时候加训了喵~