[二分+容斥原理] poj 2773 Happy 2006

时间：2014-05-07 02:37:49 收藏：0 阅读：435

题目链接：

Happy 2006

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 9131		Accepted: 3073

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

POJ Monthly--2006.03.26,static

[Submit] [Go Back] [Status] [Discuss]

题目意思：

求与m互质的第k大的数

解题思路：

二分+容斥原理

二分出要求的数i,用容斥原理求出1~i之间的与m互质的数的个数kk,如果kk<k扩大范围，如果kk>k扩大范围，如果kk=k并且gcd(i,m)=1说明刚好找到，否则继续缩小范围。

代码：

//#include<CSpreadSheet.h>

#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

#define N 1000
bool isp[N+10];
int pri[N+10],pp[N+10],cnt,cnt0;
int m,k;
ll res;

void init()  //打出1000内的质数
{
    cnt=0;
    for(int i=1;i<=N;i++)
        isp[i]=true;

    for(int i=2;i<=N;i++)
    {
        if(isp[i])
        {
            pri[++cnt]=i;
            for(int j=i*2;j<=N;j+=i)
                isp[j]=false;
        }
    }
}
void Cal(int cur) //对cur分解质因数
{
    cnt0=0;

    for(int i=1;pri[i]*pri[i]<=cur;i++)
    {
        if(!(cur%pri[i]))
        {
            pp[++cnt0]=pri[i];
            while(!(cur%pri[i]))
                cur/=pri[i];
        }
    }
    if(cur!=1)
        pp[++cnt0]=cur;
}

void dfs(ll hav,int cur,int num,ll va) //容斥原理求出与m互质的且小于va的数的个数
{
    if(cur>cnt0||(hav>va))
        return ;
    for(int i=cur;i<=cnt0;i++)
    {
        ll temp=hav*pp[i];

        if(num&1)
            res-=va/temp;
        else
            res+=va/temp;
        dfs(temp,i+1,num+1,va);
    }
}
void solve(ll cur) //求出1~cur间与m互质的数的个数
{
    res=cur;

    for(int i=1;i<=cnt0;i++)
    {
        res-=cur/pp[i];
        dfs(pp[i],i+1,2,cur);
    }
}
ll gcd(ll a,ll b) //求最大公约数
{
    if(a%b==0)
        return b;
    return gcd(b,a%b);
}
int main()
{
   //freopen("in.txt","r",stdin);
   //freopen("out.txt","w",stdout);
   init();
   while(~scanf("%d%d",&m,&k))
   {
       ll ans,l,r,mid;

       l=1,r=(1LL<<62LL); //初始化一个很大的数
       Cal(m);

       while(l<=r)
       {
           mid=(l+r)>>1;
           solve(mid);
           ll temp=res; //个数

           //printf("mid:%I64d res:%I64d\n",mid,res);
           //system("pause");

           if(temp<k) //放大
                l=mid+1;
           else if(temp==k)
           {
               if(gcd(mid,m)==1) //最后一个刚好是与m互质的，说明恰好找到
               {
                   ans=mid;
                   break;
               }
               r=mid-1; //否则继续缩小
           }
           else
                r=mid-1;
       }
       printf("%I64d\n",ans);
   }
   return 0;
}

[二分+容斥原理] poj 2773 Happy 2006,布布扣,bubuko.com