【暑训排位 #D】Gym - 100952H 动态规划DP

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

Output

1

2

5

17

题意：定义回文数为首尾数值相同，且往中间递增，问和位sum的回文数有多少个。

思路：

首先，回文中心是一个数的话，由于其两边的各自和一定相等，所以除去中间一个数，两边加起来肯定是2m，然后我们就可以枚举m，求出中间的数是sum-2*m的时候，其中一边能组成和为m的数的组合。这里就是一个简单的dp问题了。

AC代码：

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <queue>
#include<sstream>
#include <stack>
#include <set>
#include <bitset>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> PII;
const int maxn = 2e2+55;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'|ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

ll dp[maxn];

int main()
{
    ll n;
    mem(dp,0); dp[0] = 1;
    while(cin>>n&&n)
    {
        ll cnt = n / 2;
        ll cur = 0;
        ll ans = 1;
        while(cnt--)
        {
            mem(dp,0);
            dp[0] = 1;
            cur += 2;
            ll oneside = cur/2;
            ll up = n - cur;
            if(n - cur == 0) rep(i,1,n/2) rep(j,i,n/2) dp[j] += dp[j-i];
            else
            rep(i,1,up) rep(j,i,oneside) dp[j] += dp[j-i];
            ans += dp[oneside];
        }
        cout<<ans<<'\n';
    }
    return 0;
}