UVA 12063 Zeros and Ones (digital DP)

Binary numbers and their pattern of bits is always very interesting to computer programmers. The problem need to count the number of positive binary numbers that has the following properties:

• The numbers is exactly N bits wide and they have no leading zeros.
• The frequency of zeros and ones are equal.
• The numbers is multiples of K.

Input

The input file contains several test cases. The first line of the input gives you the number of test cases, T ( 1T100). Then T test Cases would follow, each of the line. The input for each test case consists of integers, n ( 1n) and K ( 0K).

Output

For each set of input print, the test Case number first. Then print the number of binary numbers that has the property that we mentioned.

Sample Input

56 36 46 226 364 2

Sample Output

Case 1:1case 2:3case 3:6case 4:1662453case 5:4.,654,283,532,552,61e,+17

illustration: Here's a table showing the possible numbers for some of the sample test cases:

6 3	6 4	6 2
101010	111000	111000
	110100	110100
	101100	101100
		110010
		101010
		100110

Main topic:

That is, give you a binary sequence of length n, and then let you find out the number of sequences that can be divisible by K, the standard DP

Problem Solving Ideas:

Begins a definition of a state: Dp[i][j][k] represents the number of sequences that have a binary sequence consisting of I 0 and J 1 and the remainder of K after a K mod

Initial state: Dp[0][1][1%k]=1;

State transition equation: If we join a 1, dp[i][j+1][(k*2+1)%k]+=dp[i][j][k];

If we join a 0, dp[i+1][j][(k*2)%K] + + dp[i][j][k];

Code:

1# include<cstdio>2# include<iostream>3# include<cstring>4 5 using namespacestd;6 7typedefLong LongLL;8 9# define MAX MaxTen  OneLL Dp[max][max][max];//Dp[i][j][k] Represents a number composed of I 0,j 1, divided by the number of K of the remainder of K A  -  -  the intMainvoid) - { -     intIcase =1; -     intT;cin>>T; +      while(t-- ) -     { +Memset (DP,0,sizeof(DP)); A LL n,k; atCin>>n>>K; -printf"Case %d:", icase++); -LL NUM1 = n/2; -LL num2 =NUM1; -         if(n%2==1|| k==0 ) -         { incout<<0<<Endl; -             Continue; to         } +dp[0][1][1%K] =1; -          for(LL i =0; I <= num1;i++ ) the         { *              for(LL j =0; J <= num2;j++ ) $             {Panax Notoginseng                  for(LL L =0; l <= K1; l++ ) -                 { thedp[i+1][j][(l*2)%k]+=Dp[i][j][l]; +dp[i][j+1[(l*2+1)%k]+=Dp[i][j][l]; A                 } the             } +         } -cout<<dp[num1][num2][0]<<Endl; $     } $  -     return 0; -}

UVA 12063 Zeros and Ones (digital DP)