Saturday, November 28, 2015

worms eat leaves|Uneaten Leaves – HackerEarth - Zenefits


https://crypticarticles.wordpress.com/2015/08/10/uneaten-leaves-hackerearth-interview-question/
K caterpillars are eating their way through N leaves. Each caterpillar falls from leaf to leaf in a unique sequence.All caterpillars start at a twig in position 0 and fall onto the leaves at positions between 1 and N. Each caterpillar i has an associated ‘jump-number’ A . A caterpillar with jump number j eats leaves at positions that are multiples of j. It will proceed in the order j, 2j, 3j, … till it reaches the end of the leaves, then it stops and builds its cocoon.
Given a set A of K elements, K ≤ 15, N ≤ 10 , we need to determine the number of uneaten leaves.
Input Format:
N = number of leaves
K = number of caterpillars
A = Array of integer jump numbers (one per line).
Output Format:
An integer denoting the number of uneaten leaves.
Sample Input:
10
3

2

4

5

Sample Output:

4
Explanation
[2,4,5] is the 3-member set of jump numbers. All leaves which are multiples of 2, 4, and 5 are eaten. Leaves 1,3,7,9 are left, and they number 4.
    public static void main(String[] args) {
        // Read Input
        Scanner scanner = new Scanner(System.in);
        System.out.println("Input Total Number of Leaves");
        int N = scanner.nextInt();
        int balanceN = N;
        System.out.println("Enter total number of Caterpillars");
        int K = scanner.nextInt();
        int[] A = new int[K];
 
        System.out.println("Array of integer jump numbers (one per line).");
        for (int i = 0; i < K; i++) {
            A[i]=scanner.nextInt();
            int gcd=A[i];
            for (int j = 0; j <=i && balanceN>0; j++) {
                gcd = findGCD(A[j],A[i]);
                if (gcd>1&& gcd<A[i]){
                    break;
                }
            }
            if (gcd==A[i]){
                balanceN=balanceN-(N/A[i]);
                HashSet<Integer> commonMultiples = new HashSet<Integer>();
                for (int j = 0; j <i ; j++) {
                   int lcm = lcm(A[j],A[i]);
 
                     while (lcm<=N){
                        commonMultiples.add(lcm);
                        lcm+=lcm;
                         
                    }
                }
                balanceN+=commonMultiples.size();
            }
 
        }
        System.out.println("Balance " + balanceN);
 
 
    }
 
    private static int lcm(int a, int b)
    {
        return a * (b / findGCD (a, b));
    }
 
    private static int findGCD(int number1, int number2) {
    //base case
        if(number2 == 0)
        {
            return number1;
        }
 
        return findGCD(number2, number1%number2);
    }

http://www.1point3acres.com/bbs/thread-145290-1-1.html
第二题就是虫吃叶子的题 给定一个int作为叶子的数量,再给一个数组,其中每个数代表一个虫行走的步数, 求没有被吃掉的叶子数量。 比如n = 100, bug = {2,5}, 那么一百片叶子中号码是2或者5的倍数的叶子全部被吃掉了。

  1. int numberOfUneaten(int n, int[] p){
  2.                 // find non divisible numbers in p
  3.                 Arrays.sort(p);
  4.                 for (int i = 0; i < p.length; i++){
  5.                         for (int j = 0; j < i; j++){-google 1point3acres
  6.                                 if (p[j] != 0 && p[i] % p[j] == 0){
  7.                                         p[i] = 0;. from: 1point3acres.com/bbs 
  8.                                         break;
  9.                                 }
  10.                         }
  11.                 }
  12.                 // find lcm of numbers in p
  13.                 int l = 1;
  14.                 for (int i = 0; i < p.length; i++){. from: 1point3acres.com/bbs 
  15.                         if (p[i] != 0){
  16.                                 l = lcm(l, p[i]);
  17.                         }
  18.                 }
  19.                 int res = n % l;
  20.                 // find undivisible numbers from 1 to min(lcm, n)
  21.                 int[] mark = new int[Math.min(l, n) + 1];. 鐣欏鐢宠璁哄潧-涓€浜╀笁鍒嗗湴
  22.                 for (int i = 0; i < p.length; i++){. Waral 鍗氬鏈夋洿澶氭枃绔�,
  23.                         if (p[i] != 0){. 鐣欏鐢宠璁哄潧-涓€浜╀笁鍒嗗湴
  24.                                 for (int j = p[i]; j <= Math.min(l, n); j += p[i]){
  25.                                         mark[j] = 1;
  26.                                 }. Waral 鍗氬鏈夋洿澶氭枃绔�,
  27.                         }
  28.                 }
  29.                 int count = 0;
  30.                 int count1 = 0;
  31.                 for (int i = 1; i < Math.min(l, n); i++){
  32.                         if (mark[i] == 0){
  33.                                 count++;. 鐗涗汉浜戦泦,涓€浜╀笁鍒嗗湴
  34.                                 if (i <= res) count1++;. Waral 鍗氬鏈夋洿澶氭枃绔�,
  35.                         }
  36.                 }.1point3acres缃�
  37.                 . 1point 3acres 璁哄潧
  38.                 // find undivisible numbers from 1 to n
  39.                 if (n < l) return count;. 1point 3acres 璁哄潧
  40.                 else{
  41.                         return n / l * count + count1;
  42.                 }
  43.         }.鏈枃鍘熷垱鑷�1point3acres璁哄潧
  44.         int lcm(int a, int b){
  45.                 return a * b / gcd(a, b);
  46.         }
  47.         int gcd(int a, int b){. 1point 3acres 璁哄潧
  48.                 if (a < b) return gcd(b, a);
  49.                 return b == 0 ? a : gcd(b, a % b);. visit 1point3acres.com for more.
  50.         }
http://www.1point3acres.com/bbs/thread-133522-1-1.html
此题一开始用暴力法超时,后来用最小公因数最大公倍数的方法还是有几个case超时, 最后实在不想写就直接提交了。自己觉得这题代码写得很烂
        public static int eat2(int n, int[] b) {
                HashSet<Integer> set = new HashSet<Integer>();
                for (int x : b) {
                        boolean test = false;
                        for (int y : set) { 鏉ユ簮涓€浜�.涓夊垎鍦拌鍧�. 
                                if (x % y == 0){
                                        test = true;
                                        break;
                                }
                                if (y % x == 0){
                                        test = true;
                                        set.remove(y);
                                        set.add(x);
                                        break;
                                }
                        }
                        if (!test) set.add(x);. more info on 1point3acres.com
                }
                
                int cm = 1;
                while (true) {
                        boolean test = true;
                        for (int x : set) {
                                if (x > cm || cm % x != 0) {
                                        test = false;
                                        break;
                                }
                        }
                        if (test) break;. visit 1point3acres.com for more.
                        cm++;
                }-google 1point3acres
                int count = 0;
                for (int i = 1; i <= cm; i++) {. more info on 1point3acres.com
                        boolean test = false;
                        for (int x : set) {. 鐗涗汉浜戦泦,涓€浜╀笁鍒嗗湴
                                if (i >= x && (i % x == 0)) {
                                        test = true;
                                        break;
                                }                
                        }
                        if (test) count++;
                }
                int time = n / cm;
                int rest = n % cm;. 鐣欏鐢宠璁哄潧-涓€浜╀笁鍒嗗湴
                int sum = time * count;
                
                for (int i = 1; i <= rest; i++) {. 鐣欏鐢宠璁哄潧-涓€浜╀笁鍒嗗湴
                        boolean test = false;
                        for (int x : set) {
                                if (i >= x && i % x == 0) {
                                        test = true;
                                        break;. 鍥磋鎴戜滑@1point 3 acres
                                }
. 鐣欏鐢宠璁哄潧-涓€浜╀笁鍒嗗湴
                        }
                        if (test) sum++; 鏉ユ簮涓€浜�.涓夊垎鍦拌鍧�. 
                }
                return n - sum;
        }
https://github.com/xieqilu/Qilu-leetcode/blob/master/B227.UneatenLeaves.cs
public static int UneatenLeaves(int[] a, int num){
  int res = 0;
  int numOfSubsets = 1<<a.Length;
  for(int i=1;i<numOfSubsets;i++){
    int product=1;
    int s = CountOne(i) & 1==1? 1:-1;
    for(int j=0;j<a.Length;j++){
      if(1<<j & i==1)
        product*=a[j];
    }
    res+= s*num/product;
  }
  return num-res;
}

private static int CountOne(int n){
  int count=0;
  while(n>0){
    n = n & (n-1);
    count++;
  }
  return count;
}

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