hdu 4472 Count （2012 ACM-ICPC 成都现场赛）
递推，考虑到一n可以由i * j + 1组合出来，即第二层有j个含有i个元素的子树。。。然后就可以了。。
#include<algorithm>#include<iostream>#include<cstring>#include<cstdlib>#include<fstream>#include<sstream>#include<bitset>#include<vector>#include<string>#include<cstdio>#include<cmath>#include<stack>#include<queue>#include<stack>#include<map>#include<set>#define FF(i, a, b) for(int i=a; i<b; i++)#define FD(i, a, b) for(int i=a; i>=b; i--)#define REP(i, n) for(int i=0; i<n; i++)#define CLR(a, b) memset(a, b, sizeof(a))#define debug puts("**debug**")#define LL long long#define PB push_back#define SL(a) strlen(a)using namespace std;const int N = 1111;const int MOD = 1e9 + 7;LL ans[N];int main(){ int n, cas = 1, i, j; CLR(ans, 0); ans[1] = 1; for(i = 1; i < N; i ++) { for(j = 1; i * j + 1 < N; j ++) { ans[j * i + 1] += ans[i]; ans[j * i + 1] %= MOD; } } while(cin >> n) { cout << "Case " << cas ++ << ": "; cout << ans[n] << endl; }}