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  • poj 2773(容斥原理)

     容斥原理入门题吧。

    Happy 2006
    Time Limit: 3000MS   Memory Limit: 65536K
    Total Submissions: 9798   Accepted: 3341

    Description

    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
    

    Source

     
    #include <iostream>
    #include <stdio.h>
    #include <string.h>
    #include <math.h>
    #include <algorithm>
    #include <string>
    #include <queue>
    #include <stdlib.h>
    using namespace std;
    
    int m,k;
    int mark[1001000];
    int save[1001000];
    int pcnt;
    int g[100100];
    long long int sum;
    int cnt;
    
    void getprime()
    {
        //1不是素数
        for(int i=2;i<=1000000;i++)
        {
            if(mark[i]==1) continue;
            save[pcnt++]=i;
            for(int j=i;j<=1000000;j+=i)
                mark[j]=1;
        }
    }
    
    void dfs(int n,long long num,int s,long long int key)
    {
        if(n==0)
        {
            sum += key/num;
            return ;
        }
        if(s>=cnt) return ;
    
        for(int i=s;i<cnt;i++)
        {
            if(num*g[i]>key) continue;
            else dfs(n-1,num*g[i],i+1,key);
        }
    }
    
    long long int fuc(long long int x)
    {
        if(x==1) return 1;
    
        long long ans=0;
    
        int sign=0;
        for(int i=1;i<=cnt;i++)
        {
            sum=0;
            if(sign==0)
            {
                dfs(i,1,0,x);
                ans+=sum;
            }
            else
            {
                dfs(i,1,0,x);
                ans-=sum;
            }
            sign=sign^1;
        }
    
        return x-ans;//这里面应该不会出现负数吧
    }
    
    int main()
    {
        getprime();
        while(scanf("%d%d",&m,&k)!=EOF)
        {
            //然后就是分解一个数了
            cnt=0;
            for(int i=2;i<=m;i++)
            {
                int flag=0;
                while(m%i==0)
                {
                    if(flag==0)
                    {
                        g[cnt++]=i;
                    }
                    flag=1;
                    m/=i;
                }
            }
            if(m!=1) g[cnt++]=m;
    
            //然后就是容斥原理
    
            int b=0,d=1000000000;
            while(b<d)
            {
                int mid=(b+d)/2;
                int key=fuc(mid);
                if(key>=k) d=mid;
                else b=mid+1;
            }
            printf("%d
    ",b);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/chenhuan001/p/4060819.html
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