Number of occurrences in the number is K, but one is not K.

Idea: 1. Create an auxiliary array arr [32] and count the sum of 1 in the corresponding bits of the number in the array

2. if the number to be searched is not 0 on a certain bit of I, the modulo of K in arr [I] is certainly 0; otherwise, the number to be searched is 1 in this bit, S + = 1 <I

The Code is as follows:

Int find (int * a, int Len, int K) {int arr [32] = {0}; int I, j; for (I = 0; I <Len; ++ I) {unsiged int K = A [I]; --> the negative value is shifted to 1, and an error occurs. Therefore, it first becomes unsigned int for (j = 0; j <32; ++ J) {arr [J] + = (k> J) & 1); // calculate the number of 1 in an integer} int result = 0; for (j = 0; j <31; ++ J) // calculates the value {If (ARR [J] % K! = 0) Result + = (1 <j);} If (ARR [31] % K! = 0) // signed result = 0-result; return result ;}

--> Question Deformation:

1. The number of one group is 4 K + 2, and the number of one group appears only twice, and the number of others appears 4 times. Find this number --> directly call the above Algorithm

2. Two numbers in an array appear only once, and the other numbers appear twice. Find the two numbers:

1. returns the result of an exclusive number or number of arrays.

2. Find the first bit of 1 from the number of results of an exclusive or, you can see that the two numbers to be searched must be 0 and 1 in this bit,

3. Based on this bit, the array is divided into two groups. One group has 0 bits and the other has 1 bits. Then, all the values in the two groups are different or. The result is displayed.

Number of occurrences in the number is K, but one is not K.