[ZOJ 2836] Number Puzzle

Number Puzzle Time limit: 2 Seconds Memory Limit: 65536 KB

Given a list of integers (A1, A2, ..., an), and a positive integer M, please find the number of positive integers that is Not greater than M and dividable by any integer from the given list.

Input

The input contains several test cases.

For the test case, there is and lines. The first line contains n (1 <= n <=) and M (1 <= M <= 200000000), and the second line contains A1, A2,. ., an (1 <= Ai <=, for i = 1, 2, ..., N).

Output

The For each test is in the input, and the result in a is output.

Sample Input

3 2 2 3 7 3 6 2 3 7

Sample Output

1 4

Introduction to the principle of repulsion, see Code

#include <iostream>#include<cstdio>using namespacestd;#definell Long Long#defineN 10intn,m;intA[n];intgcd (ll a,ll b) {returnB?GCD (b,a%b): A;}intLCM (ll a,ll b) {returnA/GCD (A, b) *b;}////notation 1/////intsolve () {intres=0;  for(intI=1;i< (1&LT;&LT;N); i++)    {        intCnt=0; ll LCM=1;  for(intj=0; j<n;j++)        {            if(i& (1&LT;&LT;J)) cnt++,lcm=LCM (Lcm,a[j]); }        if(cnt&1) res+=m/LCM; Elseres-=m/LCM; }    returnRes;}////notation 2/////intans;voidDfsintPosintLcmintCNT) {    if(CNT) {if(cnt&1) ans+=m/LCM; Elseans-=m/LCM; }     for(inti=pos+1; i<n;i++) Dfs (I,LCM (lcm,a[i]), CNT+1);}intMain () { while(SCANF ("%d%d", &n,&m)! =EOF) {         for(intI=0; i<n;i++) scanf ("%d",&A[i]); //printf ("%d\n", Solve ());ans=0; DFS (-1,1,0); printf ("%d\n", ans); }    return 0;}

[ZOJ 2836] Number Puzzle