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 12006 22006 3

Sample Output

135

Source

,static

 

#include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              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
              
               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
               
                =k) d=mid; else b=mid+1; } printf("%d\n",b); } return 0;}