AtCoder Beginner Contest 049

A

题目大意： 判断一个字符是不是元音字母。

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#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int maxn = 2e5 + 10;
const ll mod = 1e9 + 7;
ll inv[maxn], fac[maxn]; // 分别表示逆元和阶乘
// 快速幂
ll quickPow(ll a, ll b)
{
    ll ans = 1;
    while (b)
    {
        if (b & 1)
            ans = (ans * a) % mod;
        b >>= 1;
        a = (a * a) % mod;
    }
    return ans;
}

void init()
{
    // 求阶乘
    fac[0] = 1;
    for (int i = 1; i <= maxn; i++)
    {
        fac[i] = fac[i - 1] * i % mod;
    }
    // 求逆元
    inv[maxn - 1] = quickPow(fac[maxn - 1], mod - 2);
    for (int i = maxn - 2; i >= 0; i--)
    {
        inv[i] = inv[i + 1] * (i + 1) % mod;
    }
}
ll C(int n, int m)
{
    if (m > n)
    {
        return 0;
    }
    if (m == 0)
        return 1;
    return fac[n] * inv[m] % mod * inv[n - m] % mod;
}
ll get(ll a, ll b, ll c, ll d)
{
    return C(c - a + d - b, c - a) % mod;
}
int main()
{
    char c = getchar();
    puts( c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' ? "vowel" : "consonant" );
    return 0;
}

B

题目大意： 每个字符串输出两遍。

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#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int maxn = 2e5 + 10;
const ll mod = 1e9 + 7;
ll inv[maxn], fac[maxn]; // 分别表示逆元和阶乘
// 快速幂
ll quickPow(ll a, ll b)
{
    ll ans = 1;
    while (b)
    {
        if (b & 1)
            ans = (ans * a) % mod;
        b >>= 1;
        a = (a * a) % mod;
    }
    return ans;
}

void init()
{
    // 求阶乘
    fac[0] = 1;
    for (int i = 1; i <= maxn; i++)
    {
        fac[i] = fac[i - 1] * i % mod;
    }
    // 求逆元
    inv[maxn - 1] = quickPow(fac[maxn - 1], mod - 2);
    for (int i = maxn - 2; i >= 0; i--)
    {
        inv[i] = inv[i + 1] * (i + 1) % mod;
    }
}
ll C(int n, int m)
{
    if (m > n)
    {
        return 0;
    }
    if (m == 0)
        return 1;
    return fac[n] * inv[m] % mod * inv[n - m] % mod;
}
ll get(ll a, ll b, ll c, ll d)
{
    return C(c - a + d - b, c - a) % mod;
}
int main()
{
    int n,m;
    string s;
    cin >> n >> m;
    while (n--)
    {
        cin >> s;
        cout << s << endl<< s << endl;
    }
    return 0;
}

C

题目大意： 给你一个字符串，请你判断这个字符串能否由若干个 "dream"、"dreamer"、"erase"、"eraser" 中的其中一个字符串拼接而成。

直接dp或者倒着枚举。

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#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int maxn = 2e5 + 10;
const ll mod = 1e9 + 7;
ll inv[maxn], fac[maxn]; // 分别表示逆元和阶乘
// 快速幂
ll quickPow(ll a, ll b)
{
    ll ans = 1;
    while (b)
    {
        if (b & 1)
            ans = (ans * a) % mod;
        b >>= 1;
        a = (a * a) % mod;
    }
    return ans;
}

void init()
{
    // 求阶乘
    fac[0] = 1;
    for (int i = 1; i <= maxn; i++)
    {
        fac[i] = fac[i - 1] * i % mod;
    }
    // 求逆元
    inv[maxn - 1] = quickPow(fac[maxn - 1], mod - 2);
    for (int i = maxn - 2; i >= 0; i--)
    {
        inv[i] = inv[i + 1] * (i + 1) % mod;
    }
}
ll C(int n, int m)
{
    if (m > n)
    {
        return 0;
    }
    if (m == 0)
        return 1;
    return fac[n] * inv[m] % mod * inv[n - m] % mod;
}
ll get(ll a, ll b, ll c, ll d)
{
    return C(c - a + d - b, c - a) % mod;

}
int dp[maxn];
int main()
{
    string s;
    cin >> s;
    dp[0] = 1;
    for (int i = 1; i <= s.length(); ++i)
    {
        if (i >= 5)
        {
            dp[i] |= (dp[i - 5] && s.substr(i - 5, 5) == "dream");
            dp[i] |= (dp[i - 5] && s.substr(i - 5, 5) == "erase");
        }
        if (i >= 6)
        {
            dp[i] |= (dp[i - 6] && s.substr(i - 6, 6) == "eraser");
        }
        if (i >= 7)
        {
            dp[i] |= (dp[i - 7] && s.substr(i - 7, 7) == "dreamer");
        }
    }
    if (dp[s.length()])
        puts("YES");
    else
        puts("NO");
    return 0;
}

D

题目大意： 给你 N 个城市，K 条公路，L 条铁路。你需要回答出任意一个城市，有多少城市和它公路连通，有多少城市和它铁路连通。

直接用并查集解决联通，用map和pair进行优化。

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#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int maxn = 2e5 + 10;
const ll mod = 1e9 + 7;
ll inv[maxn], fac[maxn]; // 分别表示逆元和阶乘
// 快速幂
ll quickPow(ll a, ll b)
{
    ll ans = 1;
    while (b)
    {
        if (b & 1)
            ans = (ans * a) % mod;
        b >>= 1;
        a = (a * a) % mod;
    }
    return ans;
}

void init()
{
    // 求阶乘
    fac[0] = 1;
    for (int i = 1; i <= maxn; i++)
    {
        fac[i] = fac[i - 1] * i % mod;
    }
    // 求逆元
    inv[maxn - 1] = quickPow(fac[maxn - 1], mod - 2);
    for (int i = maxn - 2; i >= 0; i--)
    {
        inv[i] = inv[i + 1] * (i + 1) % mod;
    }
}
ll C(int n, int m)
{
    if (m > n)
    {
        return 0;
    }
    if (m == 0)
        return 1;
    return fac[n] * inv[m] % mod * inv[n - m] % mod;
}
ll get(ll a, ll b, ll c, ll d)
{
    return C(c - a + d - b, c - a) % mod;
}
int n, k, l;
const int N = 1e6 + 100;
int f1[N];
int f2[N];
int find(int x, int *fa)
{
    return fa[x] == x ? x : fa[x] = find(fa[x], fa);
}
void merge(int x, int y, int *fa)
{
    fa[find(x, fa)] = fa[find(y, fa)];
}
map<pair<int,int>,int>mp;
int main()
{
    cin >> n >> k >> l;
    for (int i = 1; i <= n; i++)
    {
        f2[i] = i;
        f1[i] = i;
    }
    for (int i = 1; i <= k; i++)
    {
        int x, y;
        cin >> x >> y;
        merge(x, y, f1);
    }
    for (int i = 1; i <= l; i++)
    {
        int x, y;
        cin >> x >> y;
        merge(x, y, f2);
    }
    for (int i = 1; i <= n; i++)
    {
        mp[{find(i, f1), find(i, f2)}]++;
    }
    for (int i = 1; i <= n; i++)
    {
        cout << mp[{find(i, f1), find(i, f2)}] << ' ';
    }
    cout << '\n';
    return 0;
}