POJ 1221: UNIMODAL PALINDROMIC DECOMPOSITIONS

Description

 

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. See the example on the next page. 

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

The basic idea is to establish a look-up table (LUT) and utilize previous calculated results. So we have to find a way to reduce a problem (decompose m) into a simpler or already calculated problem.

For an odd number, we can either leave it untouched to get an answer, or decompose it into 3 or more components. For example, we can decompose 17 into 1 + 15 + 1. By limiting both ends of the sequence to 1, we can now focus on finding the number of ways to decompose 15. In this way, we reduce the current problem to a smaller one, and we can do it recursively. We can also limit the ends of the sequence to 3, but when decomposing the middle part, both ends cannot be less than 3. 

For an even number, there is an extra way to perform decomposition, i.e. dividing the number by half. For example, 12 = 6 + 6.

In my code below, I establish an LUT to store the intermediate results. I hope the code is easier to understand than my explanation above. Anyway, it's my first blog and I hope it could help someone in need. 

 

#include <iostream>
#include <cstring>

using namespace std;

const int maxn = 251;

long long d[maxn][maxn];

int main()
{
    memset(d, 0, sizeof(d));
    d[1][1] = 1;
    d[2][1] = 1;
    d[2][2] = 1;
    d[1][0] = 1;
    d[2][0] = 2; 
    for(int i = 3; i < maxn; i++)
    {
        d[i][i] = 1;
        d[i][0] += 1;
        if(i % 2 == 0)
        {
            d[i][i/2] = 1;
            d[i][0] += 1;
        }
        for(int j = 1; j < maxn && i >= 3 * j; j++)
        {
            for(int m = j; m <= i - 2 * j; m++)
            {
                d[i][j] += d[i - 2 * j][m];
            }
            d[i][0] += d[i][j];
        }
    }

    int n;
    cin >> n;
    while(n)
    {
        cout << n << ' ' << d[n][0] << endl;
        cin >> n;
    }
    return 0;
}