A Fifty-Fifty

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#include<bits/stdc++.h>
using namespace std;
#define int long long
int read()
{
    int x = 0, f = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
        if(ch == '-') f = -1;
        ch = getchar();
    } 
    while(ch >= '0' && ch <= '9')
    {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}

signed main()
{
	vector<int> a(26);
    string s;
    cin >> s;
    for(int i = 0; i < 4; i++) a[s[i] - 'A']++;
    for(int i = 0; i < 26; i++)
    {
        if(a[i] && a[i] != 2)
        {
            cout << "No\n";
            return 0;
        }
    }
    cout << "Yes\n";
	return 0;
}

B Ordinary Number

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#include<bits/stdc++.h>
using namespace std;
#define int long long
int read()
{
    int x = 0, f = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
        if(ch == '-') f = -1;
        ch = getchar();
    } 
    while(ch >= '0' && ch <= '9')
    {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}

signed main()
{
	int n = read();
    vector<int> a(n);
    for(int i = 0; i < n; i++) a[i] = read();
    int res = 0;
    for(int i = 0; i < n - 2; i++)
    {
        if(a[i] > a[i + 1] && a[i + 1] > a[i + 2])
            res++;
        if(a[i] < a[i + 1] && a[i + 1] < a[i + 2])
            res++;
    }
    cout << res;
	return 0;
}

C Divide the Problems

找到第n / 2 个和第n / 2 + 1个数求差

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#include<bits/stdc++.h>
using namespace std;
#define int long long
int read()
{
    int x = 0, f = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
        if(ch == '-') f = -1;
        ch = getchar();
    } 
    while(ch >= '0' && ch <= '9')
    {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}

signed main()
{
	int n = read();
    vector<int> a(n);
    for(int i = 0; i < n; i++) a[i] = read();
    sort(a.begin(), a.end());
    cout << a[n / 2] - a[n / 2 - 1];
	return 0;
}

D Blue and Red Balls

排列组合问题

类似与隔板法， 如果要移动$i$次，则要有$i$组Blue Balls插入在Red Balls之间(或者两边)，选择插入的位置共有$C_{n-k +1}^{i}$

接下来就是如何将Blue Balls分成$i$组，总共又$k-1$个可分隔的地方，从中选出$i-1$个就分成了$i$组，即$C_{k-1}^{i-1}$

所以移动$i$的总方案数就是$ C_{k-1}^{i-1} * C_{n-k +1}^{i} $

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#include<bits/stdc++.h>
using namespace std;
#define int long long
int read()
{
    int x = 0, f = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
        if(ch == '-') f = -1;
        ch = getchar();
    } 
    while(ch >= '0' && ch <= '9')
    {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}

const int mod = 1e9 + 7;

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



signed main()
{
	int n = read(), k = read();
    vector<int> f(n + 1), inv(n + 1);
    f[0] = 1;
    for(int i = 1; i <= n; i++) f[i] = f[i - 1] * i % mod;
    inv[0] = 1;
    inv[n] = qmi(f[n], mod - 2);
    for(int i = n - 1; i >= 1; i--) inv[i] = inv[i + 1] * (i + 1) % mod;
    auto C = [&](int a, int b) -> int {
        if(b < a) return 0;
        return f[b] * inv[b - a] % mod * inv[a] % mod;
    };
    for(int i = 1; i <= k; i++) cout << (C(i - 1, k - 1) * C (i, n - k + 1)) % mod << endl;
	return 0;
}

E Hopscotch Addict

分层图最短路，主要是理解题意，然后建模

因为要是三的倍数所以分成三层，建图对于一条边就 1-> 2 -> 3-> 1 这样

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#include<bits/stdc++.h>
using namespace std;
#define int long long
int read()
{
    int x = 0, f = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
        if(ch == '-') f = -1;
        ch = getchar();
    } 
    while(ch >= '0' && ch <= '9')
    {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}

typedef pair<int, int> PII;

signed main()
{
	int n = read(), m = read();
    vector<vector<int>> a(3 * n + 1);
    for(int i = 0; i < m; i++)
    {
        int u = read(), v = read();
        a[u].emplace_back(v + n);
        a[u + n].emplace_back(v + 2 * n);
        a[u + 2 * n].emplace_back(v);
    }
    int s = read(), t = read();
    vector<int> dist(3 * n + 1, 0x3f3f3f3f);
    vector<int> st(3 * n + 1);
    auto bfs = [&]() -> void {
        queue<int> q;
        q.push(s);st[s] = 1;dist[s] = 0;
        while(q.size())
        {
            auto ver = q.front();
            q.pop();
            for(auto x : a[ver])
            {
                if(st[x]) continue;
                dist[x] = dist[ver] + 1;
                st[x] = 1;
                q.push(x);
            }
        }
    };
    bfs();
    if(dist[t] == 0x3f3f3f3f) cout << -1;
    else cout << dist[t] / 3;
	return 0;
}

F Hopscotch Addict

有点晚了，看了下题感觉不是我能写的，有缘再见，看题解是数据结构优化dp。。。