There is an array with n elements a1, a2, …, an and the number x.In one operation you can select some i (1 ≤ i ≤ n) and replace element ai with ai & x, where & denotes the bitwise and operation.You want the array to have at least two equal elements after applying some operations (possibly, none). In other words, there should be at least two distinct indices i ≠ j such that ai = aj. Determine whether it is possible to achieve and, if possible, the minimal number of operations to apply.
The first line contains integers n and x (2 ≤ n ≤ 100 000, 1 ≤ x ≤ 100 000), number of elements in the array and the number to and with.The second line contains n integers ai (1 ≤ ai ≤ 100 000), the elements of the array.
Print a single integer denoting the minimal number of operations to do, or -1, if it is impossible
Input 4 3 1 2 3 7 Output 1 Input 2 228 1 1 Output 0 Input 3 7 1 2 3 Output -1
In the first example one can apply the operation to the last element of the array. That replaces 7 with 3, so we achieve the goal in one move. In the second example the array already has two equal elements In the third example applying the operation won’t change the array at all, so it is impossible to make some pair of elements equal.
#include<stdio.h> int main() { long n,x,a,i,j,f=0,s,mi=-1,b[2][100001]={0}; scanf("%ld %ld",&n,&x); for(i=0;i<n;i++) { scanf("%ld",&a); b[0][a]++; s=x&a; b[1][s]++; if(a==s) b[1][s]--; } for(i=0;i<=100000;i++) { if(b[0][i]>1) { mi=0; break; } else if(b[1][i]+b[0][i]>1&&b[0][i]!=0) mi=1; else if(b[1][i]>1&&mi==-1) mi=2; } printf("%ld\n",mi); return 0; }Because direct nesting of two loops results in time overtime, the data is not read and stored directly, but the number of values per value is stored through an array first. Analysis knows that there is at most one value that is different from the original data after the operation, so take an array to store the number of values after the bit.
