UVa 1645 Count (递推，数论)

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;

}

巴特西

UVa 1645 Count (递推，数论)

最新文章

热门文章