codility: Euclidean algorithm ( ChocolatesByNumbers, CommonPrimeDivisors)

Chocolates By Numbers

Two positive integers N and M are given. Integer N represents the number of chocolates arranged in a circle, numbered from 0 to N − 1.

You start to eat the chocolates. After eating a chocolate you leave only a wrapper.

You begin with eating chocolate number 0. Then you omit the next M − 1 chocolates or wrappers on the circle, and eat the following one.

More precisely, if you ate chocolate number X, then you will next eat the chocolate with number (X + M) modulo N (remainder of division).

You stop eating when you encounter an empty wrapper.

For example, given integers N = 10 and M = 4. You will eat the following chocolates: 0, 4, 8, 2, 6.

The goal is to count the number of chocolates that you will eat, following the above rules.

Write a function:

int solution(int N, int M);

that, given two positive integers N and M, returns the number of chocolates that you will eat.

For example, given integers N = 10 and M = 4. the function should return 5, as explained above.

Assume that:

• N and M are integers within the range [1..1,000,000,000].

Complexity:

• expected worst-case time complexity is O(log(N+M));
• expected worst-case space complexity is O(1).

不知道怎么证明N/gcd(N,M)，从直观上其实比较好理解。。

// you can also use includes, for example:
// #include <algorithm>
int gcd (int a, int b) {
    if(a<b) {
        return gcd(b,a);
    }
    if(a%b==0) {
        return b;
    }
    return gcd(b,a%b);
}
int solution(int N, int M) {
    // write your code in C++98
    return N/gcd(N,M);
}

Common Prime Divisors:

A prime is a positive integer X that has exactly two distinct divisors: 1 and X. The first few prime integers are 2, 3, 5, 7, 11 and 13.

A prime D is called a prime divisor of a positive integer P if there exists a positive integer K such that D * K = P. For example, 2 and 5 are prime divisors of 20.

You are given two positive integers N and M. The goal is to check whether the sets of prime divisors of integers N and M are exactly the same.

For example, given:

• N = 15 and M = 75, the prime divisors are the same: {3, 5};
• N = 10 and M = 30, the prime divisors aren't the same: {2, 5} is not equal to {2, 3, 5};
• N = 9 and M = 5, the prime divisors aren't the same: {3} is not equal to {5}.

Write a function:

int solution(vector<int> &A, vector<int> &B);

that, given two non-empty zero-indexed arrays A and B of Z integers, returns the number of positions K for which the prime divisors of A[K] and B[K] are exactly the same.

For example, given:

    A[0] = 15   B[0] = 75
    A[1] = 10   B[1] = 30
    A[2] = 3    B[2] = 5

the function should return 1, because only one pair (15, 75) has the same set of prime divisors.

Assume that:

• Z is an integer within the range [1..6,000];
• each element of arrays A, B is an integer within the range [1..2147483647].

Complexity:

• expected worst-case time complexity is O(Z*log(max(A)+max(B))2);
• expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).

这个比较好办，因为如果a和b是由相同的prime divisor组成，那么a/gcd(a,b)必然是gcd(a,b)的一个因子，对b同理。

// you can also use includes, for example:
#include <algorithm>
int gcd(int a, int b) {
    if(a<b) {
        return gcd(b,a);
    }
    if(a%b==0) {
        return b;
    }
    return gcd(b,a%b);
}
bool isCommon(int a, int b) {
    int gcdValue = gcd(a,b);
    int gcdA,gcdB;
    while(a!=1) {
        gcdA = gcd(a,gcdValue);
        if(gcdA==1)
            break;
        a = a/gcdA;
    }
    if(a!=1)  {
        return false;
    }
    while(b!=1) {
        gcdB = gcd(b,gcdValue);
        if(gcdB==1)
            break;
        b = b/gcdB;
    }
    return b==1;
}
int solution(vector<int> &A, vector<int> &B) {
    // write your code in C++98
    int res = 0;
    for(int i=0;i<A.size();i++) {
        if(isCommon(A[i],B[i])) {
            res++;
        }
    }
    return res;
}