UNIMODAL PALINDROMIC DECOMPOSITIONS——動態規劃

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.

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

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

樣例輸出

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

提示

N < 250

DP：用f[i][j]表示左邊爲j的時候，把i分解爲單調回文數串的種類數

狀態轉移方程f[i][j] = sum{f[i-2*j][k]----->k>=j && i-2*j >= k}

注意用long long

#include<iostream>
#include<cstring>
using namespace std;
long long f[255][255];
int main()
{
	int N;
	memset(f,0,sizeof(f));
	for(int i = 1; i <= 250; ++i)
	{
		f[i][i] = 1;
		if(i % 2 == 0)
			f[i][i/2] = 1;
		for(int j = 1; i-2*j >= j; ++j)  
		{
			for(int k = j; i-2*j>=k;++k)
				f[i][j] += f[i-2*j][k];
		}
	}
   while(cin >> N)
   {
	   if(N == 0)
		   return 0;
	   long long sum = 0;
	   for(int i = 1; i <= N; ++i)
	   sum += f[N][i];
	   cout << N << ' ' << sum << endl;
   }
   system("pause");
   return 0;
}