51Nod 1315 legal integer set

A set S of integers is valid, meaning that any subset subS of S has Fun(SubS)! =X, where X is a fixed integer, Fun(A) is defined as follows:

A is a set of integers. Suppose there are n elements in A, which are a0, a1, a2,..., an-1, then define: Fun(A)=a0 or a1 or... or an-1; Fun({}) = 0, that is, the function value of the empty set is 0. Among them, or is the or operation.

Now give you a set Y and integer X values, ask at least how many elements can be removed from set Y to make set Y valid?

 

For example: Y = {1,2,4}, X=7; obviously the current Y is not legal because 1 or 2 or 4 = 7, but after removing any element, Y will be legal. So, the answer is 1.

Input

The first line contains two integers N, X, where N is the number of elements in the Y set, X is as described in the title, and 1<=N<=50, 1<=X<=1,000,000,000.
After N lines, each line contains an integer yi, that is, the ith element in the set Y, and 1<=yi<=1,000,000,000.

Output

An integer representing the minimum number of elements to delete.

Input example

5 7
1
2
4
7
8

Output example

2 
Note that it is an arbitrary subset, then first find out all the elements that may get x, and finally find the contribution value of each element

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#pragma comment(linker, "/stck:1024000000,1024000000")
#define lowbit(x) (x&(-x))
#define max(x,y) (x>=y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.1415926535897932384626433832
#define ios() ios::sync_with_stdio(true)
#define INF 0x3f3f3f3f
#define mem(a) ((a,0,sizeof(a)))
typedef long long ll;
ll a[55],b[55],res[55],n,x,ans,y,cnt=100;
int top=0;
int main()
{
    scanf("%lld%lld",&n,&x);
    years = x;
    while (years)
    {
        a[top ++]=ans% 2 ;
        years /= 2 ;
    }
    for(int i=0;i<n;i++)
    {
        scanf("%lld",&y);
        if((y|x)!=x) continue;
        int top=0;
        while(y)
        {
            if(y&1) b[top]++;
            and /= 2 ;
            top++;
        }
    }
    for(int i=0;i<=50;i++)
        if(a[i]) cnt=min(cnt,b[i]);
    printf("%d\n",cnt);
    return 0;
}