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Happy 2006(poj2773)

Happy 2006

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 10856		Accepted: 3776

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
思路：容斥原理+二分；
先打表素数，然后求要求的数的不同质因数；然后二分找数，用容斥求在[1,mid]与N不互质的数，然后求得与N互质的数，
最后最小那个[1,mid]有m个与N互质的数的mid即所求；

  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<iostream>
  4 #include<string.h>
  5 #include<queue>
  6 #include<stack>
  7 #include<map>
  8 using namespace std;
  9 typedef  long long LL;
 10 bool prime[1000005];
 11 LL ans[100005];
 12 LL fen[100];
 13 LL ak[12];
 14 LL cc[3000];
 15 queue<LL>que;
 16 int main(void)
 17 {
 18         LL n,m;
 19         int i,j;
 20         for(i=2; i<2000; i++)
 21         {
 22                 if(!prime[i])
 23                 {
 24                         for(j=i; i*j<=1000000; j++)
 25                         {
 26                                 prime[i*j]=true;
 27                         }
 28                 }
 29         }
 30         int cnt=0;
 31         for(i=2; i<=1000000; i++)
 32         {
 33                 if(!prime[i])
 34                 {
 35                         ans[cnt++]=(LL)i;
 36                 }
 37         }
 38         while(scanf("%lld %lld",&n,&m)!=EOF)
 39         {
 40 
 41                 {
 42                         LL nn=n;
 43                         int flag=0;
 44                         int cn=0;
 45                         int t=0;
 46                         while(nn>1)
 47                         {
 48                                 if(!flag&&nn%ans[t]==0)
 49                                 {
 50                                         flag=1;
 51                                         nn/=ans[t];
 52                                         que.push(ans[t]);
 53                                 }
 54                                 else if(nn%ans[t]==0&&flag)
 55                                 {
 56                                         nn/=ans[t];
 57                                 }
 58                                 else
 59                                 {
 60                                         flag=0;
 61                                         t++;
 62                                 }
 63                         }
 64                         while(!que.empty())
 65                         {
 66                                 fen[cn]=que.front();
 67                                 que.pop();
 68                                 cn++;
 69                         }
 70                         for(i=1; i<=(1<<cn)-1; i++)
 71                         {
 72                                 int  t=0;
 73                                 LL cp=1;
 74                                 for(j=0; j<cn; j++)
 75                                 {
 76                                         if(i&(1<<j))
 77                                         {
 78                                                 t++;
 79                                                 cp*=fen[j];
 80                                         }
 81                                 }
 82                                 if(t%2)
 83                                 {
 84                                         cc[i]=cp;
 85                                 }
 86                                 else cc[i]=-cp;
 87                         }
 88                         LL l;
 89                         LL r;
 90                         l=1;
 91                         r=1e18;
 92                         LL id=0;
 93                         while(l<=r)
 94                         {
 95                                 LL mid=(l+r)/2;
 96                                 LL sum=0;
 97                                 for(i=1; i<=(1<<cn)-1; i++)
 98                                 {
 99                                         sum=sum+(mid/cc[i]);
100                                 }
101                                 sum=mid-sum;
102                                 if(sum>=m)
103                                 {
104                                         id=mid;
105                                         r=mid-1;
106                                 }
107                                 else
108                                 {
109                                         l=mid+1;
110                                 }
111                         }
112                         printf("%lld
",id);
113                 }
114         }
115         return 0;
116 }

油！油！you@

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原文地址：https://www.cnblogs.com/zzuli2sjy/p/5465698.html