AtCoder Beginner Contest 044(4/4)

news2025/1/17 15:50:35

Tak and Hotels (ABC Edit)

前k晚花费x，k+1晚以后花费y

AC代码：

import java.util.*;
import java.io.*;
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Solve Work = new Solve();
        int T = 1;
        // int T = Integer.parseInt(in.next());
        for (int i = 1; i <= T; i++) {
            Work.main(i, in, out);
        }
        out.close();
    }
    static class Solve {
        public void main(int testNumber, InputReader in, PrintWriter out) {
            int n = in.nextInt(), k = in.nextInt(), x = in.nextInt(), y = in.nextInt();
            int ans = 0;
            if (n >= k) {
                ans = k * x + (n - k) * y;
            } else {
                ans = n * x;
            }
            out.println(ans);
        }
    }
    //IO
    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;
        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }
        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
        public int nextInt() {
            return Integer.parseInt(next());
        }
        public long nextLong() {
            return Long.parseLong(next());
        }
        public double nextDouble() {
            return Double.parseDouble(next());
        }
    }
}

Beautiful Strings

字符串中出现的字符，每种字符的个数为偶数

AC代码：

import java.util.*;
import java.io.*;
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Solve Work = new Solve();
        int T = 1;
        // int T = Integer.parseInt(in.next());
        for (int i = 1; i <= T; i++) {
            Work.main(i, in, out);
        }
        out.close();
    }
    static class Solve {
        public void main(int testNumber, InputReader in, PrintWriter out) {
            String s = in.next();
            int len = s.length();
            int[] cnt = new int[26];
            for (int i = 0; i < len; i++) {
                cnt[s.charAt(i) - 'a']++;
            }
            boolean ok = true;
            for (int i = 0; i < 26; i++) {
                if (cnt[i] % 2 == 1) {
                    ok = false;
                    break;
                }
            }
            if (ok) {
                out.println("Yes");
            } else {
                out.println("No");
            }
        }
    }
    //IO
    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;
        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }
        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
        public int nextInt() {
            return Integer.parseInt(next());
        }
        public long nextLong() {
            return Long.parseLong(next());
        }
        public double nextDouble() {
            return Double.parseDouble(next());
        }
    }
}

Tak and Cards

题意是在序列中选择任意个数使得他们的平均数为A的方案数

DP，考虑到尽量不要使用浮点数，那么就将A扩大对应的倍数即可，显然是个01背包问题，这里直接用滚动数组优化后的dp，状态就是前i个里面选j个和为k，其中j<=i，k<= $\text{[math]}$ ,因为滚动数组中存的是上一轮状态的解，所以这里要从大到小枚举，可得转移方程dp[j][k]=dp[j][k]+dp[j-1][k-x[i]]

AC代码：

import java.util.*;
import java.io.*;
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Solve Work = new Solve();
        int T = 1;
        // int T = Integer.parseInt(in.next());
        for (int i = 1; i <= T; i++) {
            Work.main(i, in, out);
        }
        out.close();
    }
    static class Solve {
        public void main(int testNumber, InputReader in, PrintWriter out) {
            int n = in.nextInt(), A = in.nextInt();
            int[] x = new int[n + 1];
            for (int i = 1; i <= n; i++) {
                x[i] = in.nextInt();
            }
            long[][] dp = new long[(n + 1)][500 * 500 + 10];
            dp[0][0] = 1;
            int sum = 0;
            for (int i = 1; i <= n; i++) {
                sum += x[i];
                for (int j = i; j >= 1; j--) {
                    for (int k = sum; k >= x[i]; k--) {
                        dp[j][k] = dp[j][k] + dp[j - 1][k - x[i]];
                    }
                }
            }
            long ans = 0;
            for (int i = 1; i <= n; i++) {
                ans = ans + dp[i][i * A];
            }
            out.println(ans);
        }
    }
    //IO
    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;
        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }
        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
        public int nextInt() {
            return Integer.parseInt(next());
        }
        public long nextLong() {
            return Long.parseLong(next());
        }
        public double nextDouble() {
            return Double.parseDouble(next());
        }
    }
}

Digit Sum

求最小的b进制，使得在b进制下的n的各位之和为s，如果不存在输出-1

可以观察到当n==s的时候答案为n+1，当n<s时答案为-1

最后讨论当n>s时

$\text{[math]}$

当b的最高幂次大于等于2的时候，

$\text{[math]}$

进而

$\text{[math]}$

所以可以暴力枚举b从[2, $\text{[math]}$ ]，判断是否有满足条件的最小的b

当b的最高幂次小于2的时候，

$\text{[math]}$

联立上述两式可得

$\text{[math]}$

有了上述公式，可以从大到小枚举 $\text{[math]}$ ，这样相当于从小到大在枚举b，从而可以找到最小的符合条件的b，由于前面已经筛过了b从[2, $\text{[math]}$ ]，这里只需要枚举b从( $\text{[math]}$ ,n-s]，并且可以得到 $\text{[math]}$ ，所以只需要枚举 $\text{[math]}$ 的部分即可

AC代码：

import java.util.*;
import java.io.*;
import java.math.*;
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Solve Work = new Solve();
        int T = 1;
        // int T = Integer.parseInt(in.next());
        for (int i = 1; i <= T; i++) {
            Work.main(i, in, out);
        }
        out.close();
    }
    static class Solve {
        public void main(int testNumber, InputReader in, PrintWriter out) {
            long n = in.nextLong(), s = in.nextLong();
            long z = (long)Math.sqrt(1.0 * n) + 1;
            for (long i = 2; i <= z; i++) {
                if (check(i, n, s)) {
                    out.println(i);
                    return;
                }
            }
            if (n < s) {
                out.println(-1);
                return;
            }
            if (n == s) {
                out.println(n + 1);
                return;
            }
            long x = n - s, y = x / z + 1;
            for (long i = y; i >= 1; i--) {
                if (x % i == 0 && s - i >= 0 && s - i < x / i + 1 && i < x / i + 1) {
                    out.println(x / i + 1);
                    return;
                }
            }
            out.println(-1);
        }
        public boolean check(long base, long n, long s) {
            long res = 0;
            while (n > 0) {
                res += n % base;
                n /= base;
            }
            if (res == s) {
                return true;
            } else {
                return false;
            }
        }
    }
    //IO
    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;
        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }
        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
        public int nextInt() {
            return Integer.parseInt(next());
        }
        public long nextLong() {
            return Long.parseLong(next());
        }
        public double nextDouble() {
            return Double.parseDouble(next());
        }
    }
}

发现一个按照D改编的原题jjgg的难题

AC代码：

#include <bits/stdc++.h>
using namespace std;
using LL = long long;
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    LL n, s;
    cin >> n >> s;
    if (n < s) {
        cout << "-1\n";
        exit(0);
    }
    if (n == s) {
        cout << n + 1 << '\n';
        exit(0);
    }
    LL ans = 1999999999999LL;
    LL p = n - s;
    function<LL(LL)> check = [&](LL z) {
        z++;
        LL sum = 0, x = n;
        while (x) {
            sum += x % z;
            x /= z;
        }
        return sum;
    };
    for (LL i = 1; i * i <= p; i++) {
        if (p % i == 0) {
            if (check(i) == s) {
                ans = min(ans, i + 1);
            }
            if (check(p / i) == s) {
                ans = min(ans, p / i + 1);
            }
        }
    }
    if (ans != 1999999999999LL) {
        cout << ans << '\n';
    } else {
        cout << "-1\n";
    }
    return 0;
}

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