UVA - 1645 - Count（思路）

題意：輸入n（1 <= n <= 1000），輸出有n個結點且每一個深度中全部結點的子節點數相同的樹有多少種。

根據題意，其實要求每一個子樹都相同。

一個結點看成根節點，還剩下n - 1個結點，枚舉n - 1的因子（因子看成緊鄰根結點的子樹中的結點數），而後將全部因子的答案相加便可。

代碼以下：

 

#include<cstdio>
#include<cstring>
#include<cctype>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<deque>
#include<queue>
#include<stack>
#include<list>
#define fin freopen("in.txt", "r", stdin)
#define fout freopen("out.txt", "w", stdout)
#define pr(x) cout << #x << " : " << x << "   "
#define prln(x) cout << #x << " : " << x << endl
#define Min(a, b) a < b ? a : b
#define Max(a, b) a < b ? b : a
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const double pi = acos(-1.0);
const double EPS = 1e-6;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const ll MOD = 1e9 + 7;
using namespace std;

#define NDEBUG
#include<cassert>
const int MAXN = 1000 + 10;
const int MAXT = 10000 + 10;

ll c[MAXN];
void init(){
    c[0] = 0ll;  c[1] = 1ll;  c[2] = 1ll;
    for(int i = 3; i < MAXN; ++i){
        int t = i - 1;
        c[i] = 0ll;
        for(int j = 1; j <= t; ++j)
            if(t % j == 0)  (c[i] += c[j]) %= MOD;
    }
}

int main(){
    init();
    int kase = 0, n;
    while(scanf("%d", &n) == 1)
        printf("Case %d: %lld\n", ++kase, c[n]);
    return 0;
}