題目地址
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

題目大意：F（n） 就是 數字總數等于n，為回文字符串，前面的子串遞增，后面的遞減的字符串的數量（看一下題目應該能懂）。
思路：由于是回文字符串，所有我們只要c處理一邊就行了，數字是對半的。如果字符串是長度是偶數，那n的數值必須是偶數，因為兩邊的總數要相等。我們先dp，預處理出整數劃分，f[i][j]表示只從1~i中選，且總和等于j的方案數。然后，我們分字符長度為偶數和奇數處理就行了。

代碼:

#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)//長度為偶數的時候，一邊的總數必定是n/2；{for(int i=1;i<=n/2;i++) ans+=f[n/2][i];}for(int i=n;i>=1;i-=2)//奇數時，枚舉中間那個數的值{for(int j=0;j<=i;j++) //因為中間的數是最大的，不得超過中間的數{//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();} }

整數劃分模板：
狀態轉移方程:
f[i][j] = f[i - 1][j] + f[i][j - i];

const int N = 1010;//一維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;// 二維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; }

總結

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