POJ1211

UNIMODAL PALINDROMIC DECOMPOSITIONS

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4279		Accepted: 2105

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

Source

Greater New York 2002

 

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

long long dp[301][301];
int n;

int main()
{
    while(scanf("%d",&n)!=EOF && n)
    {
        memset(dp,0,sizeof(dp));
        for(int j=0; j<=n; j++)
        {
            dp[0][j]=1;
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=i/2+1;j<=i;j++)
            {
                dp[i][j]=1;
            }
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=i/2;j>=1;j--)
            {
                dp[i][j]=dp[i-2*j][j]+dp[i][j+1];
            }
        }
        printf("%d %lld\n",n,dp[n][1]);
    }
    return 0;
}