HUAWEI Programming Contest 2024（AtCoder Beginner Contest 342）

HUAWEI Programming Contest 2024（AtCoder Beginner Contest 342）

A - Yay!

代码:

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
#define fi first
#define se second
using i128 = __int128_t;
using piii = pair<ll, pair<ll, ll>>;

void solve()
{
    string s;
    cin >> s;
    int n = s.size();
    vector<int> cnt(30), pos(30);
    for (int i = 0; i < n; i++)
    {
        cnt[s[i] - 'a']++;
        pos[s[i] - 'a'] = i + 1;
    }
    for (int i = 0; i < 26; i++)
    {
        if (cnt[i] == 1)
        {
            cout << pos[i] << endl;
            return;
        }
    }
}

int main()
{
    int t = 1;
    // cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

B - Which is ahead?

代码:

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
#define fi first
#define se second
using i128 = __int128_t;
using piii = pair<ll, pair<ll, ll>>;

void solve()
{
    int n;
    cin >> n;
    vector<int> a(n + 1), pos(110);
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
        pos[a[i]] = i;
    }
    int q;
    cin >> q;
    while (q--)
    {
        int l, r;
        cin >> l >> r;
        if (pos[l] < pos[r])
        {
            cout << l << endl;
        }
        else
        {
            cout << r << endl;
        }
    }
}

int main()
{
    int t = 1;
    // cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

C - Many Replacement

解题思路:

每次变换，\(26\)个字母一个个看。

代码:

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
#define fi first
#define se second
using i128 = __int128_t;
using piii = pair<ll, pair<ll, ll>>;

void solve()
{
    int n;
    cin >> n;
    string s;
    cin >> s;
    int q;
    cin >> q;
    vector<char> f(40);
    for (int i = 0; i < 26; i++)
    {
        char c = i + 'a';
        f[i] = c;
    }
    while (q--)
    {
        char a, b;
        cin >> a >> b;
        for (int i = 0; i < 26; i++)
        {
            if (f[i] == a)
            {
                f[i] = b;
            }
        }
    }
    for (auto c : s)
    {
        cout << f[c - 'a'];
    }
    cout << endl;
}

int main()
{
    int t = 1;
    // cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

D - Square Pair

解题思路:

\(cnt[i]:表示在序列A中， 指数为奇数的质因子的乘积为i，这样的数字的个数。\)

平方数的质因子指数都是偶数。

对于两个平方数而言，相乘还是平方数。他们指数为奇数的质因子个数为零。

对于两个非平方数，只有他们指数为奇数的质因子相乘起来，这些质因子指数变为偶数才会变为平方数。由于奇数加奇数一定为偶数，我们不用记录指数具体是多少，只要记录其奇偶性即可。

最后计数时，\(0\)不止和自己可以组合，它还可以和任何其他数组合。

代码:

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
#define fi first
#define se second
using i128 = __int128_t;
using piii = pair<ll, pair<ll, ll>>;

void solve()
{
    int n;
    cin >> n;
    vector<ll> a(n + 1);
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
    }
    vector<ll> primes, minp(2e5 + 10);
    for (int i = 2; i <= 2e5; i++)
    {
        if (!minp[i])
        {
            minp[i] = i;
            primes.push_back(i);
        }
        for (auto p : primes)
        {
            if (i * p > 2e5)
            {
                break;
            }
            minp[i * p] = p;
            if (i % p == 0)
            {
                break;
            }
        }
    }
    vector<ll> cnt(2e5 + 10);
    // 平方数的质因子指数都是偶数。
    for (int i = 1; i <= n; i++)
    {
        if (a[i] == 0)
        {
            cnt[a[i]]++;
            continue;
        }
        // 只留下指数为奇数的质因子，奇数加奇数为偶数
        for (auto p : primes)
        {
            if (p * p > a[i])
            {
                break;
            }
            while (a[i] % (p * p) == 0)
            {
                a[i] /= p * p;
            }
        }
        cnt[a[i]]++;
    }
    ll ans = 0;
    ll res = 0;
    for (int i = 1; i <= 2e5; i++)
    {
        ans += cnt[i] * (cnt[i] - 1) / 2;
        res += cnt[i];
    }
    ans += cnt[0] * (cnt[0] - 1) / 2;
    // 这里特例，0和任何数相乘都是平方数
    ans += cnt[0] * res;
    cout << ans << endl;
}

int main()
{
    int t = 1;
    // cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

E - Last Train

解题思路:

从点\(N\)开始跑最长路。

代码:

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
#define fi first
#define se second
using i128 = __int128_t;
using piii = pair<ll, pair<ll, ll>>;
const ll inf = 1ll << 60;
void solve()
{
    int n, m;
    cin >> n >> m;
    vector<vector<array<ll, 5>>> e(n + 10, vector<array<ll, 5>>());
    for (int i = 1; i <= m; i++)
    {
        ll l, d, k, c, a, b;
        cin >> l >> d >> k >> c >> a >> b;
        e[b].push_back({a, l, d, k, c});
    }
    vector<ll> dist(n + 10, -1);
    priority_queue<pii> q;
    q.push({inf, n});
    while (q.size())
    {
        auto [ds, u] = q.top();
        q.pop();
        if (dist[u] != -1)
        {
            continue;
        }
        dist[u] = ds;
        for (auto [v, l, d, k, c] : e[u])
        {
            // 减去路途时间
            ll t = ds - c;
            if (t >= l)
            {
                // 最晚出发时间
                t = l + min((k - 1), (t - l) / d) * d;
                q.emplace(t, v);
            }
        }
    }
    for (int i = 1; i < n; i++)
    {
        if (dist[i] == -1)
        {
            puts("Unreachable");
        }
        else
        {
            cout << dist[i] << "\n";
        }
    }
}

int main()
{
    int t = 1;
    // cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}