Educational Codeforces Round 47

problem1009
JAVA

A. GameShopping

签到题,不多说

    public static void main(String[] args) throws InterruptedException {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt(), m = scan.nextInt();
        int[] cost = new int[n];
        Queue<Integer> q = new LinkedBlockingQueue<>();
        for (int i = 0; i < n; i++) {
            cost[i] = scan.nextInt();
        }
        for (int i = 0; i < m; i++) {
            ((LinkedBlockingQueue<Integer>) q).put(scan.nextInt());
        }
        int ans = 0;
        Integer head = null;
        for (int i = 0; i < n; i++) {
            head = q.peek();
            if (null == head) {
                break;
            }
            if (head >= cost[i]) {
                q.poll();
                ans++;
            }
        }
        System.out.println(ans);
    }

B. Minimum Ternary String

根据题意可知1可以随意移动,2与0无法移动. 所以可以很快推出2与0的相对位置不会变化,1会通通移动到2前面. 那么只要把第一个2前的0全部先输出,然后输出1,最后再按顺序输出的2与0即可

    public static void countWay(String s) {
        int zeroBeforeTwo = 0;
        int oneNum = 0;
        boolean twoFlag = false;
        Queue<Character> q = new LinkedBlockingQueue<>();
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (c == '1') {
                oneNum++;
                continue;
            } else if (c == '2') {
                twoFlag = true;
            }
            if (twoFlag) {
                q.add(c);
            } else if (c == '0') {
                zeroBeforeTwo++;
            }
        }
        //开始生成新字符
        char[] newChar = new char[s.length()];
        int p = 0;
        for (int i = 0; i < zeroBeforeTwo; i++) {
            newChar[p++] = '0';
        }
        for (int i = 0; i < oneNum; i++) {
            newChar[p++] = '1';
        }
        while (!q.isEmpty()) {
            newChar[p++] = q.poll();
        }
        System.out.println(new String(newChar));
    }

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        String s = scan.nextLine();
//        stackSimulation(s);
//        twoAsSpliter(s);
        countWay(s);
    }

C. Annoying Present

题目较为简单,等差数列求和与二次函数求最值稍微推一下就可以出来. 唯一需要注意的就是精度问题,根据输入数据量可以推出longlong先加不会爆,故可先累加最后再除n

/**
 * 输入n,m 以及m个 xi,di
 * avg(变换后)-avg(原来的) = Xi+Di*( (n*n+n)/2 + i*i - (n+1)*i )/n 
 * 故后半段括号内最小值为 (n+1)/2 , 最大值为1或n
 * 仅需根据Di正负取值即可
 * 注意精度问题,n可最后再除
 */
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        long n, m, xi, di;
        n = scan.nextInt();
        m = scan.nextInt();
        long avg = 0;
        for (int i = 0; i < m; i++) {
            xi = scan.nextInt();
            di = scan.nextInt();
            avg += xi * n;
            if (0 == di) {
            } else if (di > 0) {
                avg += afterDiMultiply(n, di, 1);
            } else {
                avg += afterDiMultiply(n, di, ((n + 1) >> 1));
            }
        }
        System.out.println(String.format("%.15f", (double) avg / n));
    }
    public static long afterDiMultiply(long n, long d, long i) {
        return d * ((n * n + n) / 2 + i * (i - 1 - n));
    }

D. Relatively Prime Graph

这道题因为给的m范围太小导致暴力枚举可直接过....

	public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt(), m = scan.nextInt();
        int[][] ansArr = new int[m][2];
        int ansCount = 0;
        if (m < n - 1) {
            System.out.println("Impossible");
            return;
        }
        for (int i = 1; i <= n; i++) {
            for (int j = i + 1; j <= n; j++) {
                if (gcd(i, j) == 1) {
                    ansArr[ansCount][0] = i;
                    ansArr[ansCount++][1] = j;
                    if (ansCount >= m) {
                        System.out.println("Possible");
                        for (int k = 0; k < ansCount; k++) {
                            System.out.println(ansArr[k][0] + " " + ansArr[k][1]);
                        }
                        return;
                    }
                }
            }
        }
        System.out.println("Impossible");
    }

    public static int gcd(int a, int b) {
        return (a == 0) ? b : gcd(b % a, a);
    }