Codility上的练习 (10)

标签：des   style   blog   io   ar   color   os   sp   for   

分析： 可以证明吃巧克力必然形成一个从0号开始的圈。因为0， M % N,  M * 2 % N .... 这些编号，如果有两个相等，比如a * M % N 和 b * M % N，满足0 < a < b 且a * M % N == b * M % N, 则有（b - a) * M % N == 0 ，说明再出现b * M % N 之前，已经出现了0。 于是问题等价于 求一个最小的正整数满足 x * M % N == 0 ,这样的x = N / gcd(M , N)。

// you can also use includes, for example:
// #include <algorithm>

int gcd(int x,int y) {
    return y?gcd(y, x % y):x;
}
int solution(int N, int M) {
    // write your code in C++98
    return N / gcd(M, N);
}

// you can use includes, for example:
// #include <algorithm>

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

int gcd(int x,int y) {
    return y?gcd(y, x % y):x;
}

bool same(int x,int y) {
    for (; y > 1; ) {
        y = gcd(x, y);
        x /= y;
    }
    return (x == 1);
}

int solution(vector<int> &A, vector<int> &B) {
    // write your code in C++11
    int n = A.size(), answer = 0;
    for (int i = 0;  i < n; ++i) {
        int g = gcd(A【i】, B【i】);
        if (same(A【i】, g) && same(B【i】, g)) {
            ++answer;
        }
    }
    return answer;
    
}

// you can use includes, for example:
// #include <algorithm>

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

int mul(long long x,long long y,int m) {
    return x * y % m;
}

int powermod(int x,int y,int m) {
int r = 1 % m;
    for (; y ; y >>= 1) {
        if (y & 1) {
            r = mul(r, x, m);
        }
        x = mul(x, x, m);
    }
    return r;
}

int solution(vector<int> &A, vector<int> &B) {
    // write your code in C++11
int n = A.size(), answer = 0;
    for (int i = 0; i < n; ++i) {
        if ((powermod(A【i】, B【i】, B【i】) == 0) && (powermod(B【i】, A【i】, A【i】) == 0)) {
            ++answer;
        }
    }
    return answer;
}

Codility上的练习 (10)

标签：des   style   blog   io   ar   color   os   sp   for   

原文地址：http://blog.csdn.net/caopengcs/article/details/41819659