POJ-2773 Happy 2006(容斥，二分，dfs)

缺乏、安全感 2022-06-09 09:51 119阅读 0赞

## Happy 2006 ##

Two positive integers are said to be relatively prime to each other if the Great Common Divisor (GCD) is 1. For instance, 1, 3, 5, 7, 9…are all relatively prime to 2006.

Now your job is easy: for the given integer m, find the K-th element which is relatively prime to m when these elements are sorted in ascending order.  
**Input**  
The input contains multiple test cases. For each test case, it contains two integers m (1 <= m <= 1000000), K (1 <= K <= 100000000).  
**Output**  
Output the K-th element in a single line.  
**Sample Input**  
2006 1  
2006 2  
2006 3  
**Sample Output**  
1  
3  
5

**题意**：给你m和k，从小到大排序，问第k个与m互质的数是多少。

**思路**：首先打一个素数表，把m分解成素因数，然后我们可以通过容斥定理求出x以内与m互质的数的个数，然后我们通过二分来找x，使得find(x)=k，满足条件的x会有多个，最后循环递减找一下最小的满足条件的x就是答案。

#include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <string>
    #include <cstdlib>
    #include <cmath>
    #include <vector>
    #include <queue>
    #include <map>
    #include <algorithm>
    #include <set>
    #include <functional>
    using namespace std;
    typedef long long LL;
    typedef unsigned long long ULL;
    const int INF = 1e9 + 5;
    const int MAXN = 1000007;
    const int MOD = 30021;
    const double eps = 1e-8;
    const double PI = acos(-1.0);
    
    int prime[MAXN], primesize;
    bool isprime[MAXN];
    void getlist(int listsize)
    {
        memset(isprime, 1, sizeof(isprime));
        isprime[1] = false;
        for (int i = 2; i <= listsize; i++)
        {
            if (isprime[i])prime[++primesize] = i;
            for (int j = 1; j <= primesize&&i*prime[j] <= listsize; j++)
            {
                isprime[i*prime[j]] = false;
                if (i%prime[j] == 0)break;
            }
        }
    }
    
    int t;
    int num[10];
    
    void solve(int x)
    {
        t = 0;
        memset(num, 0, sizeof num);
        for (int i = 1; prime[i]*prime[i] <= x; i++)
        {
            if (x%prime[i] == 0)
                t++;
            while (x%prime[i]==0)
            {
                num[t] = prime[i];
                x /= prime[i];
            }
        }
        if (x != 1)
            num[++t] = x;
    }
    
    LL ans;
    
    void dfs(LL k,LL place,LL mul,LL flag)
    {
        if (place == t)
        {
            if (mul != 1)
                if (flag)ans += k / mul;
                else ans -= k / mul;
            return;
        }
        dfs(k, place + 1, mul, flag);
        dfs(k, place + 1, mul*num[place + 1], flag ^ 1);
    }
    
    LL find(LL x)
    {
        ans = 0;
        dfs(x, 0, 1, 0);
        return x - ans;
    }
    
    LL Find(int x)
    {
        LL mid;
        LL l = 1;
        LL r = 1LL << 60;
        while (l<r)
        {
            mid = (l + r) / 2;
            if (find(mid) == x)
                return mid;
            if (find(mid) > x)
                r = mid;
            else
                l = mid;
        }
    }
    
    int main()
    {
        int m, K;
        LL s;
        getlist(1000003);
        while (scanf("%d%d",&m,&K)!=EOF)
        {
            solve(m);
            s=Find(K);
            while (find(s-1)==K)
                s--;
            printf("%lld\n", s);
        }
    }