比賽地址Educational Codeforces Round 116 (Rated for Div. 2)

A. AB Balance

容易看出，直接判斷首尾字母是否相同即可，因為最後要化成 aba...a 或者 bab...b 形式才行

#include <bits/stdc++.h>
#define endl '\n'
#define ls u << 1
#define rs u << 1 | 1
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL,LL> PLL;
const int INF = 0x3f3f3f3f, N = 1e5 + 10;
const int MOD = 1e9 + 7;
const double eps = 1e-6;
const double PI = acos(-1);
inline int lowbit(int x) {return x & (-x);}

inline void solve() {
    string s; cin >> s;
    s[0] = s[s.size() - 1];
    cout << s << endl;
}
int main() {
#ifdef DEBUG
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    auto now = clock();
#endif
    ios::sync_with_stdio(false), cin.tie(nullptr);
    cout << fixed << setprecision(2);
    int T; cin >> T;
    while (T -- )
        solve();
#ifdef DEBUG
    cout << "============================" << endl;
    cout << "Program run for " << (clock() - now) / (double)CLOCKS_PER_SEC * 1000 << " ms." << endl;
#endif
    return 0;
}
 

B. Update Files

根據題意模擬即可

#include <bits/stdc++.h>
#define endl '\n'
#define ls u << 1
#define rs u << 1 | 1
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL,LL> PLL;
const int INF = 0x3f3f3f3f, N = 1e5 + 10;
const int MOD = 1e9 + 7;
const double eps = 1e-6;
const double PI = acos(-1);
inline int lowbit(int x) {return x & (-x);}

inline void solve() {
    LL n, k; cin >> n >> k;
    LL now = 1, ans = 0, tag = 1;
    while (now < n) {
        ans ++;
        now += tag;
        tag <<= 1ll;
        if (tag > k) {
            ans += (n - now + k - 1) / k;
            break;
        }
    }
    cout << ans << endl;
}
int main() {
#ifdef DEBUG
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    auto now = clock();
#endif
    ios::sync_with_stdio(false), cin.tie(nullptr);
    cout << fixed << setprecision(2);
    int T; cin >> T;
    while (T -- )
        solve();
#ifdef DEBUG
    cout << "============================" << endl;
    cout << "Program run for " << (clock() - now) / (double)CLOCKS_PER_SEC * 1000 << " ms." << endl;
#endif
    return 0;
}
 

C. Banknotes

貪心即可

容易看出，比能拼的數多出 \(1\) 即可。

#include <bits/stdc++.h>
#define endl '\n'
#define ls u << 1
#define rs u << 1 | 1
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL,LL> PLL;
const int INF = 0x3f3f3f3f, N = 1e5 + 10;
const int MOD = 1e9 + 7;
const double eps = 1e-6;
const double PI = acos(-1);
inline int lowbit(int x) {return x & (-x);}

inline void solve() {
    LL n, k; cin >> n >> k;
    vector<LL> a(n + 1);
    for (int i = 1; i <= n; i ++ ) {
        cin >> a[i];
        a[i] = (LL)pow(10, a[i]);
    }
    k ++;
    LL res = 0;
    for (int i = 1; i < n; i ++ ) {
        LL d = a[i + 1] / a[i] - 1;
        if (k >= d) {
            res += a[i] * d;
            k -= d;
        } else {
            res += a[i] * k;
            k = 0;
        }
    }
    if (!k) cout << res << endl;
    else {
        res += a[n] * k;
        cout << res << endl;
    }
}
int main() {
#ifdef DEBUG
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    auto now = clock();
#endif
    ios::sync_with_stdio(false), cin.tie(nullptr);
    cout << fixed << setprecision(2);
    int T; cin >> T;
    while (T -- )
        solve();
#ifdef DEBUG
    cout << "============================" << endl;
    cout << "Program run for " << (clock() - now) / (double)CLOCKS_PER_SEC * 1000 << " ms." << endl;
#endif
    return 0;
}
 

D. Red-Blue Matrix

待補

E. Arena

考慮dp

狀態設計 \(dp[i][j]\) 表示有 \(i\) 個人，血量範圍是 \(j\) 的方案數

首先考慮最簡單的，當每個人第一輪被淘汰的情況，容易看出只能在 \(1到n - 1\) 取值，對於 \(n\) 個人，每人有 \(n - 1\) 種取法，同時考慮到其血量範圍，故取 \(dp[n][x] = min(n - 1, x) ^ {n}\)。

如果 \(x < n - 1\) 那麼只可能是第一輪就被淘汰完，否則繼續考慮

考慮每輪的淘汰人數，由於不能存在最後勝利者，所以我們取每輪淘汰的人數為 \(k\)， \(k \in [0, n - 2]\)

然後直接轉移即可

\(dp[n][x] = \sum_{k=0}^{n - 2}(n - 1)^k * dp[n - k][x - n + 1] * C_n^i\)

稍微解釋一下，由於本輪淘汰了 \(k\) 人，並且本輪所有人扣的血量為 \(n - 1\)，所以我們只能從 \(dp[n - k][x - n + 1]\) 中轉移，相當於 \(k\) 個人可以分配到 \(1到n - 1\) 的血量，所以是 \((n - 1)^k\)，在算上從 \(n\) 個人中選擇 \(i\) 個人淘汰 \(C_n^i\)，得到轉移式。

用記憶化搜尋寫即可

#include <bits/stdc++.h>
#define endl '\n'
#define ls u << 1
#define rs u << 1 | 1
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL,LL> PLL;
const int INF = 0x3f3f3f3f, N = 550;
const int MOD = 998244353;
const double eps = 1e-6;
const double PI = acos(-1);
inline int lowbit(int x) {return x & (-x);}

LL dp[N][N], c[N][N];

LL q_pow(LL a, LL b) {
    LL res = 1;
    for (; b; b >>= 1) {
        if (b & 1) res = res * a % MOD;
        a = a * a % MOD;
    }
    return res;
}

LL dfs(int n, int x) {
    if (dp[n][x]) return dp[n][x];
    dp[n][x] = q_pow(min(x, n - 1), n);
    if (x <= n - 1) return dp[n][x];
    for (int i = 0; i <= n - 2; i ++ ) dp[n][x] = (dp[n][x] + q_pow(n - 1, i) * dfs(n - i, x - n + 1) % MOD * c[n][i] % MOD) % MOD;
    return dp[n][x];
}

inline void solve() {
    int n, x; cin >> n >> x;
    for (int i = 0; i <= 500; i ++ ) {
        for (int j = 0; j <= i; j ++ ) {
            if (i == 0 || j == 0) c[i][j] = 1;
            else c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % MOD;
        }
    }
    cout << dfs(n, x) << endl;
}
int main() {
#ifdef DEBUG
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    auto now = clock();
#endif
    ios::sync_with_stdio(false), cin.tie(nullptr);
    cout << fixed << setprecision(2);
//    int T; cin >> T;
//    while (T -- )
        solve();
#ifdef DEBUG
    cout << "============================" << endl;
    cout << "Program run for " << (clock() - now) / (double)CLOCKS_PER_SEC * 1000 << " ms." << endl;
#endif
    return 0;
}