poj1221

思路：递推or dp

f[i][j]表示要分解i且最小数字为j的情况

f[i][j] +=  f [ i-2 * j ][ j ]         ( i-2*j >= j )，

f[i][j] += 1                             ( i == 2 * j ) 

初始    f[i][i]=1

/*
  State:Accepted
  Time:2013.3.2
*/
#include <iostream>
#include <cstring>
#include <string>
#include <cstdlib>
#include <fstream>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn = 250;
long long f[maxn][maxn];
int n;

void work(){
     for (int i = 0; i <= maxn; ++i)  f[i][i] = f[i][0] = 1;
     f[1][1] = f[2][1] = f[2][2] = 1;
     for (int i = 3; i <= maxn ; ++i)
        for (int j = 1; j <= i/2 ; ++j){
           if (j * 2 == i) ++f[i][j];
           else 
             for (int k = j; k <= i - 2 * j ; ++k)
                f[i][j] += f[i - 2*j][k];
        } 
    
}

void solve(){
    long long  ans = 0;
    for (int i = 0; i <= n/2 ; ++i)
        ans += f[n][i];
    printf("%d %lld\n", n, ans);
}
int main(){
     freopen("poj1221.in","r",stdin);
     freopen("poj1221.out","w",stdout);
     work();
     scanf("%d",&n);
     while ( n ){ 
            solve();
            scanf("%d",&n);
     }
     fclose(stdin); 
     fclose(stdout);
}