AtCoder 333 A-E

李菜菜271
136 阅读4分钟

A Three Threes (打表

import java.util.*;

class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();
        if (n == 1) System.out.println(1);
        if (n == 2) System.out.println(22);
        if (n == 3) System.out.println(333);
        if (n == 4) System.out.println(4444);
        if (n == 5) System.out.println(55555);
        if (n == 6) System.out.println(666666);
        if (n == 7) System.out.println(7777777);
        if (n == 8) System.out.println(88888888);
        if (n == 9) System.out.println(999999999);
    }
}

B Pentagon (判断是否是同类型的边

import java.util.*;

class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        String s12 = sc.next();
        String t12 = sc.next();

        boolean fs12 = false; // 相邻是true
        boolean ft12 = false; // 相邻是true

        if (s12.charAt(1) == s12.charAt(0) + 1 || s12.charAt(1) == 'E' && s12.charAt(0) == 'A' || s12.charAt(1) == 'A' && s12.charAt(0) == 'E' || s12.charAt(0) == s12.charAt(1) + 1) fs12 = true;
        if (t12.charAt(1) == t12.charAt(0) + 1 || t12.charAt(1) == 'E' && t12.charAt(0) == 'A' || t12.charAt(1) == 'A' && t12.charAt(0) == 'E' || t12.charAt(0) == t12.charAt(1) + 1) ft12 = true;

        if (fs12 == ft12) System.out.println("Yes");
        else System.out.println("No");
    }
}

C Repunit Trio (三重循环暴力枚举

import java.util.*;

class Main {
    static long[] arr = new long[1000010];
    static long[] base = {1, 11, 111, 1111, 11111, 111111, 1111111, 11111111, 111111111, 1111111111, 11111111111L, 111111111111L};
    static long[] ans = new long[1000010];
    static int idx = 0;
    static int idxa = 0;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();
        for (int i = 0; i < base.length; i ++ )
            for (int j = 0; j < base.length; j++)
                for (int k = 0; k < base.length; k++)
                    arr[idx++] = base[i] + base[j] + base[k];

        Arrays.sort(arr, 0, idx);
//        for (int i = 0; i < 10; i ++ ) System.out.print(arr[i] + " ");
        for (int i = 0; i < idx; i ++ ) {
            if (i == 0) ans[idxa ++ ] = arr[i];
            else if (arr[i] != arr[i - 1]) ans[idxa ++ ] = arr[i];
        }

//        System.out.println(idxa);
//        for (int i = 0; i < 20; i ++ ) System.out.print(ans[i] + " ");

        System.out.println(ans[n - 1]);
    }
}

D Erase Leaves (找出节点1的n - 1个子树的最小节点和

import java.util.*;

class Main {
    static int[] h, e, ne, arr;
    static int idx, idxarr;
    static int n;
    static int ans = 0;

    private static void init() {
        h = new int[2 * n + 100]; e = new int[2 * n + 100]; ne = new int[2 * n + 100];
        arr = new int[2 * n + 100];
        Arrays.fill(h, -1);
    }
    private static void add(int a, int b) {
        e[idx] = b; ne[idx] = h[a]; h[a] = idx ++ ;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        // 找ver1的其中n - 1棵树的最小总和
        n = sc.nextInt();
        init();
        for (int i = 1; i <= n - 1; i ++ ) {
            int u = sc.nextInt(), v = sc.nextInt();
            add(u, v);
            add(v, u);
        }

        int d1 = 0;
        for (int i = h[1]; i != -1; i = ne[i] ) {
            int j = e[i];
            arr[idxarr ++ ] = dfs(j, 1);
            d1 ++ ;
        }

        Arrays.sort(arr, 0, idxarr);
//        System.out.println("debug -> d1 : " + d1);
        for (int i = 0; i < d1 - 1; i ++ ) ans += arr[i];
        System.out.println(ans + 1);
    }

    public static int dfs(int u, int fa) {
        int ret = 1;
        for (int i = h[u]; i != -1; i = ne[i]) {
            int j = e[i];
            if (j == fa) continue;
            ret += dfs(j, u);
        }

        return ret;
    }
}

AtCoder 333. E Takahashi Quest (贪心 + 栈维护

import java.util.*;

class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();

        /**
         * 思路 : 贪心
         * 策略 : 当我需要用伤害为x[i]的药水的时候优先使用理我最近的一天的那瓶伤害为x[i]的药水
         * 实现 : 利用栈维护
         * */

        int[] t = new int[n + 10], x = new int[n + 10];
        boolean[] f = new boolean[n + 10]; // 代表第i天的药水是否有用, false代表有用, true代表无用, 其中有用(false)有两种含义 -> 一种是打怪日, 一种是需要捡起来以后用
        boolean[] at = new boolean[n + 10]; // 查看第i天是否是打怪日

        for (int i = 1; i <= n; i ++ ) {
            t[i] = sc.nextInt();
            x[i] = sc.nextInt();
        }

        ArrayList<Stack<Integer>> ast = new ArrayList<>(); // ArrayList放的伤害为x[i]的药水的栈, 栈里放的是该次x[i]药水出现的日期
        for (int i = 0; i < n + 10; i ++ ) ast.add(new Stack<>()); // 初始化

        for (int i = 1; i <= n; i ++ ) {
            if (t[i] == 1) ast.get(x[i]).push(i); // 第i天的药水先捡起来放到x[i]的栈里
            else {
                if (ast.get(x[i]).isEmpty()) { // 如果需要用到伤害为x[i]的药水但是栈是空的, 说明打败不了该天的怪物, 输出-1后结束程序
                    System.out.println(-1);
                    return ;
                }
                ast.get(x[i]).pop(); // 贪心策略 : 用离我最近的一天的一瓶
                at[i] = true; // 第i天是打怪日
            }
        }

        for (int i = 0; i < ast.size(); i ++ ) { // 剩下捡了的药水就是没必要捡起来的药水, 清空每一个栈, 标记f为true代表无用
            Stack<Integer> st = ast.get(i);
            while (!st.isEmpty()) {
                int ti = st.pop();
                f[ti] = true; // 第ti天的药水没用
            }
        }

        int ans = 0;
        int now = 0;
        for (int i = 1; i <= n; i ++ ) {
            if (!f[i] && !at[i]) now ++ ; // 不是打怪日, 且第i天药水显示有用, 则代表第i天需要捡起来
            if (at[i]) now -- ; // 打怪日需要用一瓶库存
            ans = Math.max(ans, now);
//            System.out.println("debug -> now : " + now + " ans : " + ans);
        }

        System.out.println(ans);
        for (int i = 1; i <= n; i ++ ) {
            if (at[i]) continue; // 打怪日无需输出
            if (!f[i]) System.out.print(1 + " ");
            else System.out.print(0 + " ");
        }
    }
}

// 写了SpecialJudge
// 1 1 1 1 1 0 1 0 0 0 0 1 1 0 1 0 0 0 1 0
// 1 1 1 1 1 0 1 0 0 0 0 1 1 0 1 0 1 0 0 0
// 1 1 1 0 0 1 0 1
// 0 0 1 1 1 0 1 1
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