The Ultimate edition of the number division--Knapsack method to solve the division of various numbers

Example: NOI 7219: Complex integer Partitioning problem

Total time limit:

200ms

Memory Limit:

65536kB

Describe

The positive integer n is represented as a series of positive integers of sum, n=n1+n2+...+nk, wherein N1>=n2>=...>=nk>=1, k>=1.
This representation of a positive integer n is called a division of positive integer n.

Input

The standard input contains several sets of test data. Each set of test data is a single row of input data, including two integers n and K.
(0 < n <=, 0 < K <= N)

Output

For each set of test data, output the following three rows of data:
First line: Number of divisions of n divided into k positive integers
Second line: Number of divisions of n divided into a number of different positive integers
Line three: number of divisions of n divided into several odd positive integers

Sample input

5 2

Sample output

233

Tips

First line: 4+1, 3+2,
Second line: 5,4+1,3+2
Line three: 5,1+1+3, 1+1+1+1+1+1

#include <cstdio>#include<cstring>#include<iostream>using namespacestd;Const intn=Wuyi;intF1[n][n],f2[n],f3[n];intSum[n],jsum[n];intn,k;voidinit () {f2[0]=f3[0]=1;  for(intI=1; i<n;++i) sum[i]=i; intt=0;  for(intI=1; i<n;i+=2) jsum[++t]=i;  for(intI=1; i<n;++i) for(intj=1; j<=i;++j) {if(j==1) F1[i][j]=1; Else if(i>=j) F1[i][j]=f1[i-1][j-1]+F1[I-J][J];/*number of divisions of n divided into k positive integers*/            }     for(intI=1; i<n;++i) for(intj=n-1; j>=sum[i];--j) F2[j]+=f2[j-sum[i]];/*number of divisions of n divided into several different positive integers*/     for(intI=1; i<=t;++i)/*Note that this is T, not N.*/       for(intj=jsum[i];j<n;++j) F3[j]+=f3[j-jsum[i]];/*number of divisions of n divided into several odd positive integers*/}intMain () {init ();  while(SCANF ("%d%d", &n,&k) = =2)    {        if(n==0) printf ("0\n0\n0\n"); Elseprintf ("%d\n%d\n%d\n", F1[n][k],f2[n],f3[n]); }    return 0;}

View Code

First line: Number of divisions of n divided into k positive integers
Second line: Number of divisions of n divided into a number of different positive integers
Line three: number of divisions of n divided into several odd positive integers

Analysis: For the "N divided into K-positive integer sum of the number of divisions", can not use knapsack method, or the traditional f[n][m]=f[n-1][m-1]+f[n-m][m];

And for n divided into several different positive integers of the number of division, you can generate a positive integer sequence as an item, run 01 backpack, the title is exactly put up the number of backpacks, note f[0]=1 on it.

For the number of divisions of "N divided into a number of odd positive integers", a positive odd number sequence is generated as an item, and a full backpack or 01 backpack depends on whether there are "different numbers".

For "dividing n into a maximum number not exceeding the number of K." ", create a 1--k sequence, run 01 backpacks or a full backpack.

The Ultimate edition of the number division--Knapsack method to solve the division of various numbers