设当前的总和为i，最外层的数字为j，把最外层“剥去”后，剩下依然是满足要求的数列，且总和为i-2j，最外层的数字>=j。
即为：F(i,j)=∑F(i-2j,k)
#include <cstdio>#include <string>int N;double f[300][300];void dp ();void pt ();void print ();int main ()
...{
dp (); //pt (); //freopen ( "in.txt", "r", stdin ); while ( scanf ( "%d", &N ) && N )
...{ print ();
} return 0;
}void dp ()...{ int i, j, k; memset ( f, 0, sizeof ( f ) ); for ( i = 1; i <= 250; i ++ ) ...{ f[i][i] = 1; for ( j = 1; j <= i / 2; j ++ ) ...{ f[i][j] = 0; if ( j * 2 == i ) f[i][j] = 1; for ( k = j; k <= i - 2 * j; k ++ ) ...{ f[i][j] += f[i - 2 * j][k]; } } }}void print ()...{ int i; double ans = 0; for ( i = 1; i <= N; i ++ ) ...{ ans += f[N][i]; } printf ( "%d %.0f ", N, ans );}