题目地址
A sequence of positive and non-zero integers called palindromic if it can be read the same forward and backward, for example:

15 2 6 4 6 2 15

20 3 1 1 3 20

We have a special kind of palindromic sequences, let’s call it a special palindrome.

A palindromic sequence is a special palindrome if its values don’t decrease up to the middle value, and of course they don’t increase from the middle to the end.

The sequences above is NOT special, while the following sequences are:

1 2 3 3 7 8 7 3 3 2 1

2 10 2

1 4 13 13 4 1

Let’s define the function F(N), which represents the number of special sequences that the sum of their values is N.

For example F(7) = 5 which are : (7), (1 5 1), (2 3 2), (1 1 3 1 1), (1 1 1 1 1 1 1)

Your job is to write a program that compute the Value F(N) for given N’s.

Input
The Input consists of a sequence of lines, each line contains a positive none zero integer N less than or equal to 250. The last line contains 0 which indicates the end of the input.

Output
Print one line for each given number N, which it the value F(N).

Examples
Input
1
3
7
10
0
Output
1
2
5
17

题目大意：F（n） 就是 数字总数等于n，为回文字符串，前面的子串递增，后面的递减的字符串的数量（看一下题目应该能懂）。
思路：由于是回文字符串，所有我们只要c处理一边就行了，数字是对半的。如果字符串是长度是偶数，那n的数值必须是偶数，因为两边的总数要相等。我们先dp，预处理出整数划分，f[i][j]表示只从1~i中选，且总和等于j的方案数。然后，我们分字符长度为偶数和奇数处理就行了。

代码:

#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <queue>
#include <cstdlib>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <bitset>
#define INF 0x3f3f3f3f3f3f3f3f
#define inf 0x3f3f3f3f
#define FILL(a,b) (memset(a,b,sizeof(a)))
#define re register
#define lson rt<<1
#define rson rt<<1|1
#define lowbit(a) ((a)&-(a))
#define ios std::ios::sync_with_stdio(false);std::cin.tie(0);std::cout.tie(0);
#define fi first
#define rep(i,n) for(int i=0;(i)<(n);i++)
#define rep1(i,n) for(int i=1;(i)<=(n);i++)
#define se secondusing namespace std;
typedef long long  ll;
typedef unsigned long long  ull;
typedef pair<int,int > pii;
const ll mod=10001;
const ll N =1e6+10;
const double eps = 1e-4;
const double pi=acos(-1);
ll gcd(int a,int b){return !b?a:gcd(b,a%b);}
int dx[4]={-1,0,1,0} , dy[4] = {0,1,0,-1};
ll f[351][351];
void solve()
{f[1][1] = 1;f[0][0] = 1;for (int i=2;i<=251;i++)for (int j=1;j<=i;j++)f[i][j]=(f[i-1][j-1]+f[i-j][j]);int n;while(cin>>n&&n){ll ans=0;if(n%2==0)//长度为偶数的时候，一边的总数必定是n/2；{for(int i=1;i<=n/2;i++) ans+=f[n/2][i];}for(int i=n;i>=1;i-=2)//奇数时，枚举中间那个数的值{for(int j=0;j<=i;j++) //因为中间的数是最大的，不得超过中间的数{//cout<<f[(n-i)/2][j]<<endl;ans+=f[(n-i)/2][j];}}cout<<ans<<endl;}
}
int main()
{iosint T;//cin>>T;T=1;while(T--){solve();}
}

整数划分模板：
状态转移方程:
f[i][j] = f[i - 1][j] + f[i][j - i];

const int N = 1010;//一维int n;
int f[N];int main()
{cin >> n;f[0] = 1;for (int i = 1; i <= n; i ++ )for (int j = i; j <= n; j ++ )f[j] = (f[j] + f[j - i]) ;cout << f[n] << endl;return 0;
}

const int N = 1010;// 二维int n;
int f[N][N];int main()
{cin >> n;f[1][1] = 1;for (int i = 2; i <= n; i ++ )for (int j = 1; j <= i; j ++ )f[i][j] = (f[i - 1][j - 1] + f[i - j][j]) ;int res = 0;for (int i = 1; i <= n; i ++ ) res = (res + f[n][i]);cout << res << endl;return 0;
}