H. Special Palindrome

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

137100

output

12517

动态规划题目，求数字n可以组成多少个回文串，dp[i][j]表示值为i，以j开头的。

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 #include <algorithm>
 5 #define ll long long
 6 using namespace std;
 7 ll dp[500][500];
 8 ll ans[500];
 9 void init(){
10     dp[0][1] = dp[1][1] = dp[2][1] = 1;
11     dp[2][2] = dp[3][1] = dp[3][3] = 1;
12     ans[1] = 1; ans[2] = ans[3] = 2;
13     for(int i = 4; i <= 250; i ++){
14         ll tmp = 0;
15         for(int j = 1; j <= i/2; j ++){
16             if(j==1)dp[i][j] = ans[i-2];
17             else{
18                 ll u = i-j*2;
19                 if(u==0)dp[i][j] = 1;
20                 else{
21                     ll ret = 0;
22                     for(int k = j; k <= u; k ++){
23                         ret += dp[u][k];
24                     }
25                     dp[i][j] = ret;
26                 }
27             }
28             tmp += dp[i][j];
29         }
30         dp[i][i] = 1;
31         tmp += 1;
32         ans[i] = tmp;
33     }
34 }
35 int main(){
36     int n;
37     init();
38     while(scanf("%d",&n)&&n){
39         cout << ans[n] << endl;
40     }
41     return 0;
42 }