poj 2773 Happy 2006
Happy 2006
Time Limit: 3000MS | Memory Limit: 65536K | |
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.
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
题意:求与n互质的第k个数
性质:若a,b互质,则a+b,b互质;若a,b不互质,则a+b,b不互质
由此可以推出,与n互质的数具有周期性:
即[1,n]与n互质的数,每个+n即可得到[n+1,2n]与n互质的数,可继续推广
所以我们只需要求出第一个周期内与n互质的数
然后ans=n*周期数+第k%φ(n)个与n互质的数
所以要求出[1,n]内与n互质的所有数
有一个小处理:如果k恰好为周期数的倍数,那么取余后为0,
所以,特殊处理,将周期数-1,再在本周期内求第φ(n)个
法一:
从n*周期数+1开始枚举,判断gcd()是否等于1,直至=1的个数达到k%φ(n)个
周期数=k/φ(n),
由于多组数据,可以欧拉筛先筛出数据范围内所有的欧拉函数值,然后O(1)使用
也可以一个一个算
我只写了第一种,Memory:5380K,Time:1579ms
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<cstdio> #define N 1000001 using namespace std; int n,k,cnt,prime[N],phi[N],t,m; bool v[N],ok; void get_euler() { phi[1]=1; for(int i=2;i<=N;i++) { if(!v[i]) { v[i]=true; prime[++cnt]=i; phi[i]=i-1; } for(int j=1;j<=cnt;j++) { if(i*prime[j]>N) break; v[i*prime[j]]=true; if(i%prime[j]==0) { phi[i*prime[j]]=phi[i]*prime[j]; break; } phi[i*prime[j]]=phi[i]*(prime[j]-1); } } } int gcd(int a,int b) { return !b ? a:gcd(b,a%b); } int main() { get_euler(); while(scanf("%d%d",&n,&k)!=EOF) { if(n==1) { printf("%d\n",k); continue; } ok=false; m=k/phi[n]; t=k-phi[n]*m; if(!t) m--,t=phi[n]; for(int i=n*m+1;;i++) { if(gcd(n,i)==1) t--; if(!t) { printf("%d\n",i); break; } } } }
法二:
埃氏筛法可以保留每个数是否与n互质,
所以对于每一组数据,都做一次埃氏筛法
然后从1开始枚举,枚举到第k%φ(n)个与n互质的数i,
ans=i+周期数*n
Memory:1156K,Time:16ms
它比法一快,因为可以O(1)判断是否与n互质,法一要gcd()判断
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<cstdio> #include<cmath> #include<cstring> using namespace std; bool check[1000001]; int euler(int n)//埃式筛法模板 { int m=int(sqrt(n+0.5)); int ans=n,k=n; memset(check,0,sizeof(check)); for(int i=2;i<=m;i++) if(n%i==0) { ans=ans/i*(i-1); for(int j=1;i*j<=k;j++) check[i*j]=true; while(n%i==0) n/=i; } if(n>1) { ans=ans/n*(n-1); for(int j=1;n*j<=k;j++) check[n*j]=true; } return ans; } int main() { int m,k,ans,cnt,t,i; while(scanf("%d%d",&m,&k)!=EOF) { ans=euler(m); cnt=0; if(k%ans==0) t=k/ans-1; else t=k/ans; k=k-ans*t; for(i=1;i<=m;i++) { if(!check[i]) cnt++; if(cnt==k) break; } printf("%d\n",i+m*t); } }