POJ 2773 Happy 2006

Happy 2006http://poj.org/problem?id=2773

Time limit:3000ms		Memory limit:65536k

Description

The positive integers is said to being relatively prime to all other if the great Common Divisor (GCD) is 1. For instance, 1, 3, 5, 7, 9...are-relatively prime to 2006.

Now your job is easy:for the given integer m, find the k-th element which was relatively prime to m when these elements ar e sorted in ascending order.

Input

The input contains multiple test cases. For each test case, it contains 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

POJ monthly--2006.03.26,staticTest Instructions: number of K for n coprimeproperty: If a, b coprime, then a+b,b coprime; If a b does not coprime, then a+b,b does not coprimeThis can be rolled out, with the number of n coprime having periodicity:that is [1,n] and n coprime number, each +n can get [n+1,2n] and n coprime number, can continue to promoteso we just need to ask for the number of coprime with N in the first cycle.then ans=n* the number of cycles + k%φ (n) and n coprimeso all the numbers in [1,n] and n coprime are required.There is a small deal: if K happens to be a multiple of the number of cycles, then the remainder is 0,So, special treatment, the number of cycles-1, and then in this week's period to find the first φ (n)Law One:enumeration from the N period number +1 to determine if GCD () is equal to 1 until the number of =1 reaches K%φ (n)Number of Cycles =k/φ (n),because of multiple sets of data, it is possible to sift out all the Euler function values in the data range, and then O (1) to usecan also be a one countI only wrote the first one, memory:5380k,time:1579ms.

#include <cstdio>#defineN 1000001using namespacestd;intn,k,cnt,prime[n],phi[n],t,m;BOOLV[n],ok;voidGet_euler () {phi[1]=1;  for(intI=2; i<=n;i++)    {        if(!V[i]) {V[i]=true; prime[++cnt]=i; Phi[i]=i-1; }         for(intj=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); }    }}intgcdintAintb) {    return!b? A:GCD (b,a%b);}intMain () {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(inti=n*m+1;; i++)        {            if(GCD (n,i) = =1) t--; if(!t) {printf ("%d\n", i);  Break; }        }    }}

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Law II:

The coprime can keep each number with N,

So for each set of data, do an Ed sieve.

Then the enumeration starts from 1, enumerating to the number I of K%φ (n) and N coprime,

ans=i+ Number of Cycles *n

Memory:1156k,time:16ms

It is faster than the law, because can O (1) to determine whether with n coprime, law one to gcd () judgment

#include <cstdio>#include<cmath>#include<cstring>using namespacestd;BOOLcheck[1000001];intEulerintN//an Egyptian sieve template{    intm=int(Sqrt (n+0.5)); intans=n,k=N; memset (check,0,sizeof(check));  for(intI=2; i<=m;i++)     if(n%i==0) {ans=ans/i* (I-1);  for(intj=1; i*j<=k;j++) Check[i*j]=true;  while(n%i==0) n/=i; }    if(n>1) {ans=ans/n* (n1);  for(intj=1; n*j<=k;j++) Check[n*j]=true; }    returnans;}intMain () {intM,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; Elset=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); }}

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POJ 2773 Happy 2006