A. Letter Song ~ 致十年后的我们

处理出来年份，\(+10\) 后输出

code:

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#include <bits/stdc++.h>

using namespace std;

string s;

int main()
{
    cin >> s;

    int x = 0;
    for (int i = 0; i < 4; i++) x = x * 10 + s[i] - '0';

    x += 10;

    cout << x;
    for (int i = 4; i < s.size(); i++) cout << s[i];

    return 0;
}

B. 简单图形问题 - 0123

手动模拟可以发现，边长为整数的等边三角形面积不可能是整数，因此只需要判断 \(s\) 是否是平方数即可

code:

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#include <bits/stdc++.h>

#define int long long

using namespace std;

int T;
int s;

signed main()
{
    cin >> T;
    while (T--)
    {
        cin >> s;
        
        int l = 1, r = s;
        while (l < r)
        {
            int mid = l + r >> 1;
            if (mid * mid < s) l = mid + 1;
            else r = mid;
        }
        int res = l * l == s? 0 : 3;
        
        cout << res << endl;
    }
    
    return 0;
}

C. 小红的机器人构造 - Ⅰ+Ⅱ+Ⅲ

组合数学

code:

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#include <bits/stdc++.h>

#define int long long

using namespace std;

const int N = 100010, mod = 1e9 + 7;

int T;
int n, x, y;
string s;
int f[N], g[N];

int qmi(int a, int b)
{
    int res = 1;
    while (b)
    {
        if (b & 1) res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}

void init()
{
    f[0] = g[0] = 1;
    for (int i = 1; i < N; i++)
    {
        f[i] = f[i - 1] * i % mod;
        g[i] = g[i - 1] * qmi(i, mod - 2) % mod;
    }
}

int C(int a, int b)
{
    if (a < b) return 0;
    return f[a] * g[b] % mod * g[a - b] % mod;
}

signed main()
{
    init();

    cin >> T;
    while (T--)
    {
        cin >> n >> x >> y;
        cin >> s;
        
        string t;
        int a = 0, b = 0, c = 0, d = 0, c1 = 0, c2 = 0;
        for (int i = 0; i < n; i++)
        {
            if (s[i] == 'U') 
            {
                a++;
                if (x > 0 && c1 < x) t += s[i], c1++;
            }
            if (s[i] == 'D') 
            {
                b++;
                if (x < 0 && c1 > x) t += s[i], c1--;
            }
            if (s[i] == 'L') 
            {
                c++;
                if (y < 0 && c2 > y) t += s[i], c2--;
            }
            if (s[i] == 'R') 
            {
                d++;
                if (y > 0 && c2 < y) t += s[i], c2++;
            }
        }
        
        if (c1 != x || c2 != y)
        {
            cout << "NO" << endl;
            continue;
        }
        
        int ca = 0, cb = 0, cc = 0, cd = 0;
        if (x > 0) ca = x;
        if (x < 0) cb = -x;
        if (y < 0) cc = -y;
        if (y > 0) cd = y;
        
        int r1 = 0, r2 = 0;
        for (int i = 0; i <= min(a - ca, b - cb); i++) 
            r1 = (r1 + C(a, i + ca) * C(b, i + cb) % mod) % mod;
        for (int i = 0; i <= min(c - cc, d - cd); i++) 
            r2 = (r2 + C(c, i + cc) * C(d, i + cd) % mod) % mod;	
        
        int res = r1 % mod * r2 % mod;
        
        cout << "YES" << ' ' << t << ' ' << res << endl;
        }

    return 0;
}

D. 小红的差值构造 - easy+hard+extreme

手玩可以发现，对于 \(x,y\) 的构造，只有两种情况：

  • \(l>n\)，不难发现，\(x\) 在 \([1,n]\) 中间位置时，\(\sum_{i=1}^{n}f_i\) 的值最小
  • \(l<=n\)，不论 \(x,y\) 取何值，\(\sum_{i=x+1}^{y-1}f_i\) 的值都是不变的，因此只需考虑两边，显然，当\([1,x]\)，与 \([y+1,n]\) 的长度接近时，答案最小

code:

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#include <bits/stdc++.h>

#define int long long

using namespace std;

int T;
int n, l;

signed main()
{
    cin >> T;
    while (T--)
    {
        cin >> n >> l;
        
        int res, x, y;
        if (l > n)
        {
            x = (n + 1) / 2, y = x + l;
            res = x * (x - 1);
            if (n % 2 == 0) res += x;
        }
        else
        {
            int t = l / 2;
            res = t * (t + 1);
            if ((l - 1) % 2) res -= t;
            x = (n - l + 1) / 2, y = x + l;
            if (x >= 1) res += x * (x - 1) / 2;
            if (y <= n) res += (n - y) * (n - y - 1) / 2;
        }
        
        cout << x << ' ' << y << ' ' << res << endl;
    }

    return 0;
}