C language dynamic planning-Sumsets (Hdu 2709)

C language dynamic planning-Sumsets (Hdu 2709)

 

Problem Description Farmer John commanded his cows to search for different sets of numbers that sum to a given number. the cows use only numbers that are an integer power of 2. here are the possible sets of numbers that sum to 7:

1) 1 + 1 + 1 + 1 + 1 + 1 + 1
2) 1 + 1 + 1 + 1 + 1 + 2
3) 1 + 1 + 1 + 2 + 2
4) 1 + 1 + 1 + 4
5) 1 + 2 + 2 + 2
6) 1 + 2 + 4

Help FJ count all possible representations for a given integer N (1 <=n <= 1,000,000 ). input A single line with a single integer, N. output The number of ways to represent N as the indicated sum. due to the potential huge size of this number, print only last 9 digits (in base 10 representation ). sample Input

7

Sample Output

6

 

Question:

Enter an integer to divide the number into an indefinite positive number and only the sum. The number must be the k power of 2 (k is a positive number greater than or equal to 0 ). number of output points. (when the output integer is too large, only the last 9 digits are output)

 

Thought 1:

 

SetA [n]ForNNumber of types;
According to the question, if n is an odd number, it is equal to the previous number.N-1Number of TypesA [n-1], IfNIs there an even number of time-based adders?1Discussion, that is, the key isNFor an even number:
1.NIs odd,A [n] = a [n-1]
2.NIs an even number:
(1) If the addition contains 1, there must be at least two 1, that isN-2After each addition+ 1 + 1, The total number of classes isA [N-2];
(2) If there is no 1 in the numberN/2Multiplied by 2, the total number of classes isA [n/2];
Therefore, the total number of types is as follows:A [n] = a [N-2] + a [n/2];

Code:

 

# Include
 
  
_ Int64 a [1000001]; int main () {_ int64 n; int I; a [1] = 1; a [2] = 2; for (I = 3; I <1000001; I ++) {if (I % 2 = 0) a [I] = a [I-2] + a [I/2]; else a [I] = a [I-1]; a [I] % = 1000000000; // The maximum number of digits is 9 .} while (scanf ("% I64d", & n )! = EOF) printf ("% I64d \ n", a [n]); return 0 ;}
 

 

Thought 2:

DP ideology,

If only 1 can be used, then the number of points for each number is 1.

If this time can be used2Is equal to or greater2NumberNYou canN-2And2Structure is convertedN-2IsD [n] = d [N-2] + d [n](FrontD [N-2]RepresentationNYou can2The number of components, followedD [n]Indicates that n can only be composed1Number of components .)

Then the state transition equation formula comes out (C [n] = 2 ^ n)

D [n] [k] = d [n] [k-1] + d [n-c [k] [k];

Cycle Dimension Reduction:

D [n] = d [n] + d [n-c [k];

 

Code:

 

#include
 
    #include 
  
   __int64 d[1000005],c[25],n,i,j;  int main()  {      while(scanf("%d",&n)!=EOF)     {        memset(d,0,sizeof(d));        c[0]=d[0]=1;      for(i=1;i<=24;i++)          c[i]=c[i-1]<<1;      for(i=0;i<=24&&c[i]<=n;i++)          for(j=c[i];j<=n;j++)              d[j]=(d[j]+d[j-c[i]])%1000000000;          printf("%d\n",d[n]);    }            return 0;}