Fjnu 1153 fat brother and xor (fat Brother with different or)

Fjnu 1153 fat brother and xor (fat Brother with different or)

Time limit:1000ms Memory limit:257792k

"Description"

"Title description"

Fat brother had master ACM, recently he began to study the operation of XOR (the operation "^"). He thought of a very interesting question:select arbitrary k positive integers from the n positive integers and then XOR the Selected K Digital, sum all cases of XOR. Now he wants to quickly calculate the results. Maybe the results would be great, just modulo 20162333. For example, 3 integers:1 2 3, select arbitrary 2 positive integers, the all cases: {2, 3}, {1, 3}. So the results is {(1 ^ 2) + (2 ^ 3) + (1 ^ 3)}%20162333

Fat elder brother is an ACM Daniel, recently he is learning XOR or operation (operator is "^"). He came up with an interesting question: choose K from n positive integers, and then add to the number of K or all the different results. Now he wants to calculate the result quickly. The end result can be very large, so modulo 20162333. For example, 3 integers: 1 2 3, select any 2 positive integers, in all cases: {2, 3}, {1, 3}. So the result is {(1 ^ 2) + (2 ^ 3) + (1 ^ 3)}%20162333

"Input"	Input
There is multiple test cases. The first line of input contains an integer t (t <=) indicating the number of test cases. For each test case: The first line contains both integer n, K (1 <= k <= n <= 1000) The second line contains n integer ai (1 <= ai <= 1000000000)	Multiple sets of test cases. The first line is an integer t (T <= 20) that represents the number of test cases. For each test case: The first line has two integers, K (1 <= k <= n <= 1000) The second line has n integers ai (1 <= ai <= 1000000000)

"Output"	Output
For per test case, output the sum% 20162333	For each test case, the output and%20162333

"Sample Input-Enter sample"	"Sample output-Output sample"
2 3 2 1 2 3 6 2 6 6 6 6 6 6	6 0

"Hint"	Prompted
The First sample test Case: (1 ^ 2) + (2 ^ 3) + (1 ^ 3) = 6 The Second sample test Case: (6 ^ 6) * 15 = 0	The first example: (1 ^ 2) + (2 ^ 3) + (1 ^ 3) = 6 A second example: (6 ^ 6) * 15 = 0

Exercises

This problem requires a binary view of each AI

As an example:

Xai represents the X-bit of the lower AI of the binary (for convenient representation starting from 0)

(A1 ^ A2) + (A2 ^ A3) + (A1 ^ A3)

=[(0a1 ^ 0a2) + (0a2 ^ 0a3) + (0a1 ^ 0a3)]*20 +

[(1a1 ^ 1a2) + (1a2 ^ 1a3) + (1a1 ^ 1a3)]*21 +

[(2a1 ^ 2a2) + (2a2 ^ 2a3) + (2a1 ^ 2a3)]*22 +

....... ..... ................. ......... ....... ..... ....................... +

[(Xa1 ^ Xa2) + (Xa2 ^ Xa3) + (Xa1 ^ Xa3)]*2x

There's only 0 and 1 left.

We can easily know: 0 ^ 0 = 0, 1 ^ 0 = 1, 1 ^ 1 = 0

Of course, a ^ b = b ^ a

So in addition to 1 ^ 0, the other result is 0 of the situation is useless.

Therefore, the title is converted to: for binary, 1 ^ 0 The number of occurrences of this situation.

Statistics AI binary X-bits appear several times 1, save to Bit[x]

1 = 2X * Available in X-bit * Number of active cases

The number of valid cases is a combination of bit[x] 1 and n-bit[x] 0.

And only an odd number of 1, the result of the XOR can be 1 (1 ^ 1 = 0)

From this we can draw

Current number of valid cases:

Finally, we use recursive formula to find the combination number quickly.

(Of course you can think of this formula as a Yang Hui triangle is also possible:))

"Code C + +"

1#include <cstdio>2#include <cstring>3 #defineMX 10054 #defineMoD 201623335 intMain () {6     intT, N, K, I, J, W, opt, bit[ -], c[mx][mx];7      for(i =0; i < MX; ++i) {8          for(c[i][0] = C[i][i] = j =1; J < I; ++j)9C[I][J] = (C[i-1][J] + c[i-1][j-1]) %MoD;Ten     } One      while(~SCANF ("%d", &t)) { A          while(t--){ -memset (bit,0,sizeof(bit)); -scanf"%d%d", &n, &k); the              for(i =0; I < n; ++i) { -scanf"%d", &j); -                  for(w =0; J J >>=1) bit[w++] + = j &1; -             } +              for(i = opt =0; I < -; ++i) { -                  for(j =1; J <= K; J + =2){ +opt + = (1LL << i) *c[bit[i]][j]% mod*c[n-bit[i]][k-j]%MoD; AOpt%=MoD; at                 } -             } -printf"%d\n", opt); -         } -     } -     return 0; in}

1#include <cstdio>2#include <cstring>3 #defineMX 10054 #defineMoD 201623335 intRead_int () {6     intAdd = GetChar ()-'0';7     intA =GetChar ();8      while(A >='0'&& a <='9') Add = Add *Ten+ A-'0', a =GetChar ();9     returnadd;Ten } One intMain () { A     intT, N, K, I, J, W, opt, bit[ -], *C[MX]; -      for(i =0; i < MX; ++i) { -C[i] =New int[i +1]; the          for(c[i][0] = C[i][i] = j =1; J < I; ++j) -C[I][J] = (C[i-1][J] + c[i-1][j-1]) %MoD; -     } -      while(t = Read_int (), t>0){ +          while(t--){ -memset (bit,0,sizeof(bit)); +n = read_int (); K =read_int (); A              for(i =0; I < n; ++i) { at                  for(w =0, j = Read_int (); J J >>=1) bit[w++] + = j &1; -             } -              for(i = opt =0; I < -; ++i) { -                  for(j =1; J <= K; J + =2){ -                     if(Bit[i] < J | | n-bit[i] < K-J)Continue; -opt + = (1LL << i) *c[bit[i]][j]% mod*c[n-bit[i]][k-j]%MoD; inOpt%=MoD; -                 } to             } +printf"%d\n", opt); -         } the     } *     return 0; $}

pseudo • Optimization

Fjnu 1153 fat brother and xor (fat Brother with different or)