ZOJ 1353 Unimodal Palindromic Decompositions（动态规划）

Unimodal Palindromic Decompositions

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Time Limit: 2 Seconds       Memory Limit: 65536 KB

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A sequence of positive integers is Palindromic if it reads the same forward and backward. For example:

23 11 15 1 37 37 1 15 11 23

1 1 2 3 4 7 7 10 7 7 4 3 2 1 1

A Palindromic sequence is Unimodal Palindromic if the values do not decrease up to the middle value and then (since the sequence is palindromic) do not increase from the middle to the end For example, the first example sequence above is NOT Unimodal Palindromic while the second example is.

A Unimodal Palindromic sequence is a Unimodal Palindromic Decomposition of an integer N, if the sum of the integers in the sequence is N. For example, all of the Unimodal Palindromic Decompositions of the first few integers are given below:

1: (1) 
2: (2), (1 1)
3: (3), (1 1 1)
4: (4), (1 2 1), (2 2), (1 1 1 1)
5: (5), (1 3 1), (1 1 1 1 1)
6: (6), (1 4 1), (2 2 2), (1 1 2 1 1), (3 3), (1 2 2 1), ( 1 1 1 1 1 1)
7: (7), (1 5 1), (2 3 2), (1 1 3 1 1), (1 1 1 1 1 1 1)
8: (8), (1 6 1), (2 4 2), (1 1 4 1 1), (1 2 2 2 1), (1 1 1 2 1 1 1), ( 4 4), (1 3 3 1), (2 2 2 2), (1 1 2 2 1 1), (1 1 1 1 1 1 1 1)

Write a program, which computes the number of Unimodal Palindromic Decompositions of an integer.

Input

Input consists of a sequence of positive integers, one per line ending with a 0 (zero) indicating the end.


Output

For each input value except the last, the output is a line containing the input value followed by a space, then the number of Unimodal Palindromic Decompositions of the input value.

Sample Input

2
3
4
5
6
7
8
10
23
24
131
213
92
0

Sample Output

2 2
3 2
4 4
5 3
6 7
7 5
8 11
10 17
23 104
24 199
131 5010688
213 1055852590
92 331143

 

动态规划 递推

观察12的单调回文分解形式中所有开头和结尾均为2的分解：

282

22422

222222

2442

假设用d[n][i]表示将n分解成单调回文中最左边的数为i的个数，例如d[12][2]表示将12分解成单调回文中最左边的数为2的个数，它的值等于以下各项之和：

d[12-2*2][2]：已经将12分解成了2...2这种形式，所以这项表示将8分解成单调回文中最左边的数为2的个数

d[12-2*2][3]：这项表示将8分解成单调回文中最左边的数为3的个数

...

 递推公式为：d[n][i] += d[n-2*i][j] (0<i<n/3  0<j<=i)

 初始条件：d[i][i] = 1 (0<i<=n)  如果i是偶数，比如12可以表示成{6,6}，所以d[i][i/2] = 1;

View Code

 1 # include<stdio.h>
 2  # include<string.h>
 3  # define maxn 512
 4  
 5  long long  d[maxn][maxn];
 6  
 7  int main(){
 8      memset(d,0,sizeof(d));
 9      int i,j;
10      //初始化
11      for(i=1;i<maxn;i++)
12      {
13          d[i][i]=1;
14          if(i%2==0) d[i][i/2]=1;  //如果i是偶数，则可分解成i/2i/2这种形式
15      }
16      d[2][1]=1;
17      d[3][1]=1;
18      d[4][1]=2;
19      d[4][2]=1;
20      
21      for(i=5;i<maxn;i++)
22      {
23          for(j=1;j<maxn;j++)
24          {
25              if(i-2*j>=j)
26              {
27                  for(int m=j;m<=i-2*j;m++)
28                      d[i][j] += d[i-2*j][m];
29              }
30              else
31                  break;
32          }
33      }
34      int n;
35      while(scanf("%d",&n) && n)
36      {
37          long long  ans=0;
38          for(i=1;i<=n;i++)
39              ans += d[n][i];
40          printf("%d %lld\n",n,ans);
41      }
42      return 0;
43  }