题意：给定一棵 n 个结点的有根树，使得每个深度中所有结点的子结点数相同。求多棵这样的树。
析：首先这棵树是有根的，那么肯定有一个根结点，然后剩下的再看能不能再分成深度相同的子树，也就是说是不是它的约数。那么答案就有了，
我们只要去计算n-1的约数有多少棵不同的树，然后就有递推式了。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std; typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e3 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
LL dp[maxn]; void init(){
dp[1] = 1; dp[2] = 1;
for(int i = 2; i < 1000; ++i){
for(int j = 1; j <= i; ++j)
if(i % j == 0) dp[i+1] = (dp[i+1] + dp[j]) % mod;
}
} int main(){
init();
int kase = 0;
while(scanf("%d", &n) == 1){
printf("Case %d: %lld\n", ++kase, dp[n]);
}
return 0;
}