题目大意:就是有n条线,通过并联或者是串联,形成一个整体,问说有n条线,可以组成多少种。
解题思路:大白书上的例题,解法还真是高端.dp[i][j]表示说每个树德叶子节点不大于i,一共有j个叶子。f[i]=dp[i-1][i],注意n为1的时候。
#include <cstdio>
#include <cstring>
typedef long long ll;
const int N = 35;
ll f[N], dp[N][N];
ll C(ll n, ll m) {
double ans = 1;
for (ll i = 0; i < m; i++)
ans *= n - i;
for (ll i = 1; i <= m; i++)
ans /= i;
return (ll)(ans+0.5);
}
void init () {
f[1] = 1;
int n = 30;
memset(dp, 0, sizeof(dp));
for (int i = 0; i <= n; i++)
dp[i][0] = 1;
for (int i = 1; i <= n; i++) {
dp[i][1] = 1;
dp[0][i] = 0;
}
for (int i = 1; i <= n; i++) {
for (int j = 2; j <= n; j++) {
for (int p = 0; p * i <= j; p++)
dp[i][j] += dp[i-1][j-i*p] * C(f[i]+p-1, p);
}
f[i+1] = dp[i][i+1];
}
}
int main () {
init ();
int n;
while (scanf("%d", &n) == 1 && n) {
printf("%lld\n", n == 1 ? 1 : 2 * f[n]);
}
return 0;
}
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