poj 1221 UNIMODAL PALINDROMIC DECOMPOSITIONS

/*C:UNIMODAL PALINDROMIC DECOMPOSITIONS

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总时间限制:
    1000ms
内存限制:
    65536kB

描述

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

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

样例输出

    2 2
    3 2
    4 4
    5 3
    6 7
    7 5
    8 11
    10 17
    23 104
    24 199
    131 5010688
    213 1 055 852 590
    92 331143

提示
    N < 250 */

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <string>
#include <cstring>
using namespace std;



int main()
{
    int n = 0;
    long long  num[255][255] = {0};
    //memset(num, 1, sizeof(num));


    for (int i = 1; i <= 251; ++i)
        for (int j = i; j >= 0; --j)
            num[i][j] = 1;

    for (int i = 0; i <= 251; ++i) num[0][i] = 1;

    for(int i = 2 ; i <= 251; ++i)
        for(int j = i / 2; j >= 1; --j)
            num[i][j] = num[i - 2 * j][j] + num[i][j + 1];
    while(true)
    {
        scanf("%d", &n);
        if(!n) break;


        printf("%d %lld\n",n , num[n][1]);
    }
    return 0;
}