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UVA - 1645 - Count（思路）

2024-08-18 14:05:42   216人阅读

题意：输入n（1 <= n <= 1000），输出有n个结点且每个深度中所有结点的子节点数相同的树有多少种。

根据题意，其实要求每个子树都相同。

一个结点当作根节点，还剩下n - 1个结点，枚举n - 1的因子（因子当作紧邻根结点的子树中的结点数），然后将所有因子的答案相加即可。

代码如下：

 

#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;
}

 

UVA - 1645 - Count（思路）


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