UNIMODAL PALINDROMIC DECOMPOSITIONS——动态规划

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

2345678102324131213920

样例输出

2 23 24 45 36 77 58 1110 1723 10424 199131 5010688213 105585259092 331143

提示

N < 250

DP：用f[i][j]表示左边为j的时候，把i分解为单调回文数串的种类数

状态转移方程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;}