POJ 2773-happy 2006 (Euler function)

Happy 2006Time limit:MS Memory Limit:65536KB 64bit IO Format:%i64d &%i64u SubmitStatusPracticePOJ 2773Appoint Description:System Crawler (2015-04-06)

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

Test instructions: Given a two number m,k, you are asked to find the number of K and M coprime.

Idea: Because there are gcd (A, B) =gcd (A+T*B,B) for the number of conventions. Therefore, in [1,m-1] with the number of M-Mutual is the same as in [K*m+1, (k+1) *m-1] The number of interconnects is equal to Phi (m), so as long as the number of [1,m-1] and M Mutual, you can find the next interval of the use of the number of the following: Then the answer is equal to the first k% The Value p+m* (K/phi (m)) of Phi (m) and M-biotin.

#include <stdio.h> #include <math.h> #include <string.h> #include <stdlib.h> #include < iostream> #include <sstream> #include <algorithm> #include <set> #include <queue> #include <stack> #include <map>using namespace std;typedef long long ll;const int inf=0x3f3f3f3f;const double pi= ACOs (    -1.0); int prime[1000010];int Euler (int n)//filter prime number + Seek <=n with n-ary numbers {int i,j;    int m=n;//because the following n is going to become, so first save n int c=n;//return is pji[m] memset (prime,0,sizeof (prime));            for (i=2;i*i<=n;i++) {if (n%i==0) {n/=i;            for (j=1;i*j<=m;j++)//Use the filter method to mark the number of all and n prime[i*j]=1;            c=c/i* (i-1);        while (n%i==0) n=n/i;        }} if (n>1) {for (j=1;n*j<=m;j++)//When n>1, n itself is also a vegetarian factor prime[n*j]=1;    c=c/n* (n-1); } return C;}    int main () {int m,k,i;    int cnt;    int sum;    int t;        while (~SCANF ("%d%d", &m,&k)) {Cnt=euler (M);t=k/cnt;        sum=0;        if (k%cnt==0) t--;        K=k-cnt*t;            for (i=1;i<=m;i++) {if (!prime[i]) sum++;        if (sum==k) break;    } printf ("%d\n", m*t+i); } return 0;}

POJ 2773-happy 2006 (Euler function)