hdu5175---Misaki's Kiss again

最新推荐文章于 2017-07-31 22:25:34 发布

原创最新推荐文章于 2017-07-31 22:25:34 发布 · 880 阅读

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本文介绍了一个有趣的算法挑战，参与者需要帮助Misaki找出所有符合条件的M值，使得GCD(N,M)=N XOR M成立。通过分析和编程实现，文章提供了一种有效的解决方案。

Problem Description
After the Ferries Wheel, many friends hope to receive the Misaki’s kiss again,so Misaki numbers them 1,2…N−1,N,if someone’s number is M and satisfied the GCD(N,M) equals to N XOR M,he will be kissed again.

Please help Misaki to find all M(1<=M<=N).

Note that:
GCD(a,b) means the greatest common divisor of a and b.
A XOR B means A exclusive or B

Input
There are multiple test cases.

For each testcase, contains a integets N(0

/*************************************************************************
    > File Name: bc30-b.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年02月14日 星期六 19时32分44秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

vector <LL> yueshu, ans;

LL gcd (LL A, LL B)
{
    return B ? gcd (B, A % B) : A;
}

int main ()
{
    LL n;
    int icase = 1;
    while (~scanf("%I64d", &n))
    {
        yueshu.clear();
        int tmp = sqrt(n + 0.5);
        for (int i = 1; i <= tmp; ++i)
        {
            if (n % i == 0)
            {
                yueshu.push_back (i);
                if (i != n / i)
                {
                    yueshu.push_back (n / i);
                }
            }
        }
        ans.clear();
        int size = yueshu.size();
        for (int i = 0; i < size; ++i)
        {
            LL u = yueshu[i];
            if ((u ^ n) <= n && (u ^ n) != 0 && gcd (n, u ^ n) == u)
            {
                ans.push_back (u ^ n);
            }
        }
        printf("Case #%d:\n", icase++);
        size = ans.size();
        sort (ans.begin(), ans.end());
        printf("%d\n", size);
        if (size != 0)
        {
            printf("%I64d", ans[0]);
            for (int i = 1; i < size; ++i)
            {
                if (ans[i] == ans[i - 1])
                {
                    continue;
                }
                printf(" %I64d", ans[i]);
            }
        }
        printf("\n");
    }
    return 0;
}