H. Special Palindrome

time limit per test:1 second
memory limit per test:64 megabytes
input:standard input
output:standard output

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

题目链接：http://codeforces.com/gym/100952/problem/H

题意：一个从开始到中间是非递减的回文被称为特殊回文，例如1123211，定义F(N)为和为N的特殊回文的个数，如F(1)=1，即和为1的回文只有一个 就是 1，F(5)=7, (7), (1 5 1), (2 3 2), (1 1 3 1 1), (1 1 1 1 1 1 1)，求F(N),N小于等于250！

思路：当N为偶数时，分2种情况，第一种为回文的长度为奇数，那么，最中间的数 m 一定是2 4 6 8......两边的数的和为(N-m)>>1，对(N-i)>>1进行整数划分（m划分），第二种为回文长度为偶数，则回文两边的和为N>>1，对N>>1整数划分(N>>1划分)，当N为奇数的时候只有一种情况，就是回文长度为奇数，最中间的数m为1 3 5 7....划分和上面一样！

下面给出AC代码：

 #include <bits/stdc++.h>

 using namespace std;

 typedef long long ll;

 inline ll read()

 {

     ll x=,f=;

     char ch=getchar();

     while(ch<''||ch>'')

     {

         if(ch=='-')

             f=-;

         ch=getchar();

     }

     while(ch>=''&&ch<='')

     {

         x=x*+ch-'';

         ch=getchar();

     }

     return x*f;

 }

 inline void write(ll x)

 {

     if(x<)

     {

         putchar('-');

         x=-x;

     }

     if(x>)

         write(x/);

     putchar(x%+'');

 }

 ll dp[][];

 inline void funday(ll n)

 {

     for(ll i=;i<=n;i++)

     {

         for(ll j=;j<=n;j++)

         {

             if(i==||j==)

                 dp[i][j]=;

             else if(i>j)

                 dp[i][j]=dp[i-j][j]+dp[i][j-];

             else if(i==j)

                 dp[i][j]=dp[i][i-]+;

             else dp[i][j]=dp[i][i];

         }

     }

 }

 ll n,m,ans;

 int main()

 {

     funday();

     while(n=read())

     {

         if(n==)

             return ;

         ans=;

         if(n&)

         {

             for(ll i=;i<=n;i+=)

                 ans+=dp[(n-i)/][i];

             ans++;

         }

         else

         {

             for(ll i=;i<=n;i+=)

                 ans+=dp[(n-i)/][i];

             ans+=dp[n/][n/];

             ans++;

         }

         write(ans);

         printf("\n");

     }

     return ;

 }

打表解法：

分奇偶用 dfs 搞出非递减的左半边串， 然后求出这个的和 ans[sum + i]++;

对于偶数个的直接dfs， 对于奇数的则枚举mid， 然后依次dfs

 void dfseven(int k, int sum)

 {

     if(*sum > ) return;

     //cout<<"here1"<<endl;

     for(int i = k; i <= ; i++){

         if(*(sum + i) <= ) {ans[*(sum + i)]++; dfseven(i, sum + i);}

         else return;

     }

 }

 void dfsodd(int mid, int k, int sum)

 {

     if(*sum + mid > ) return;

     //cout<<"here2"<<endl;

     for(int i = k; i <= ; i++){

         if(*(sum + i) + mid <=  && i <= mid) {ans[*(sum + i) + mid]++; dfsodd(mid, i, sum + i);}

         else return;

     }

 }

然后只打了前ans[50] 及以前的， 因为后面的比较大时间不够的， 所以打出前50的表然后到数列网站 OEIS 查了一下， 还真有，??

所以把那前250个ans贴到 txt里， 然后写一个中间程序 把这些数据 转换成 printf("ans[%d] = %d;", i, val);的样子打到另一个txt文件里， 然后复杂粘贴到上去， 整理下就好了

打表完整代码：

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 using namespace std;

 typedef long long LL;

 const int maxn =  + ;

 int ans[maxn];

 //偶数个的

 void dfseven(int k, int sum)

 {

     if(*sum > ) return;

     //cout<<"here1"<<endl;

     for(int i = k; i <= ; i++){

         if(*(sum + i) <= ) {ans[*(sum + i)]++; dfseven(i, sum + i);}

         else return;

     }

 }

 void dfsodd(int mid, int k, int sum)

 {

     if(*sum + mid > ) return;

     //cout<<"here2"<<endl;

     for(int i = k; i <= ; i++){

         if(*(sum + i) + mid <=  && i <= mid) {ans[*(sum + i) + mid]++; dfsodd(mid, i, sum + i);}

         else return;

     }

 }

 int main()

 {

     #ifdef LOCAL

     freopen("a.txt", "r", stdin);

     freopen("b.txt", "w", stdout);

     #endif // LOCAL

     memset(ans, , sizeof ans);

     //ans[2]++;

     dfseven(, );

     //ans[1]++;

     for(int i = ; i <= ; i++){

         dfsodd(i, , );

     }

     for(int i = ; i <= ; i++){

         printf("ans[%d] = %d;", i, ans[i] + );

     }

 /*

     int n;

     while(scanf("%d", &n)){

         printf("%d", ans[n]);

     }

     */

     return ;

 }

最终AC代码如下：

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 using namespace std;

 typedef long long LL;

 const int maxn =  + ;

 LL ans[maxn];

 int main()

 {

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     int n;

     while(scanf("%d", &n)){

         if(n == ) break;

         printf("%I64d\n", ans[n]);

     }

     return ;

 }

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