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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Manasa and Stones
  5. Discussions

Manasa and Stones

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947 Discussions

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  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can watch the explanation here : https://youtu.be/BXNoCVP0XEE Solution 1 : 

    vector<int> stones(int n, int a, int b) {
    if(a > b){
        a = b + a;
        b = a - b;
        a = a - b;
    }
    int step = b - a;
    vector<int> result;
    for(int i = a * (n-1); i < b * (n-1); i += step) result.push_back(i);
    result.push_back(b * (n-1));
    return result;
}

    Solution 2 :

    vector<int> stones(int n, int a, int b) {
    vector<int> result;
    for(int i = min(a, b) * (n-1); i < max(a,b) * (n-1); i += abs(b - a)) result.push_back(i);
    result.push_back(max(a, b) * (n-1));
    return result;
}

  • tangjiaqing
     + 0 comments

    I think there is something wrong with the examples except for the first one. The first parameter represents non-zero stone. However, in the rest example, including the test cases, n represents total number of stones including non-zero. This is the solution to pass all of the tests.

    public static List<int> Run(int count, int diff1, int diff2)
{
	if (diff1 == diff2) return [diff1 * (count - 1)];

	var small = diff1 < diff2 ? diff1 : diff2;
	var large = diff1 < diff2 ? diff2 : diff1;

	var start = small * (count - 1);
	var end = large * (count - 1);
	var step = large - small;

	if (start == 0) start += step;

	var faces = new List<int>();

	for (var i = start; i <= end; i += step)
	{
		faces.Add(i);
	}

	return faces;
}

  • zettix1
     + 0 comments
        public static List<Integer> stones(int n, int a, int b) {
        Set<Integer> resultsSet = new HashSet<>();
        for (int i = 0; i < n; i++) {
            resultsSet.add(i * a + (n - i - 1) * b);
        }
        Integer[] resultsArray = new Integer[resultsSet.size()];
        resultsSet.toArray(resultsArray);
        Arrays.sort(resultsArray);
        return Arrays.asList(resultsArray);
    }

  • yossefibrahim7oo
     + 2 comments

    Python3 

    def stones(n, a, b):
    # Write your code here
    res=[]
    for i in range(n):
        res.append((i)*b+(n-i-1)*a)
    return sorted(set(res))

  • vbthani
     + 0 comments

    for Python3 Platform

    from itertools import combinations_with_replacement

def stones(n, a, b):
    ans = set()
    diff_perm = combinations_with_replacement((a, b), n-1)
    
    for i in diff_perm:
        ans.add(sum(i))
    
        
    return sorted(ans)

T = int(input())
while (T > 0):
    n = int(input())
    a = int(input())
    b = int(input())
    
    result = stones(n, a, b)
    
    print(*result)
    
    T -= 1