ural 1142. Relations

1142. Relations

Time limit: 1.0 second
Memory limit: 64 MB

Background

Consider a specific set of comparable objects. Between two objects a and b, there exits one of the following three classified relations:

a = b
a < b
b < a

Because relation '=' is symmetric, it is not repeated above.

So, with 3 objects (a, b, c), there can exist 13 classified relations:

a = b = c       a = b < c       c < a = b       a < b = c
b = c < a       a = c < b       b < a = c       a < b < c
a < c < b       b < a < c       b < c < a       c < a < b
c < b < a

Problem

Given N, determine the number of different classified relations between N objects.

Input

Includes many integers N (in the range from 2 to 10), each number on one line. Ends with −1.

Output

For each N of input, print the number of classified relations found, each number on one line.

Sample

input	output
2 3 -1	3 13

 

Tags: none  (hide tags for unsolved problems)

Difficulty: 144

 

题意：计算N个数的大小比较关系的可能情况的总数。

分析：n个数，中间插入一些小于号。其余位置用等号。

等号相连的数的位置没有区别。

问题相当于问将n个数分为几个部分有多少种办法。

dp即可

dp[i][j]表示i个数，分为j组的方案数。

dp[i][j] += dp[i - 1][j] * j   表示第i个数有j中选择加入。

dp[i][j] += dp[i-1][j-1] * j  表示第i个数自立门户，然后这个新的部分插入其他j-1个已经确立大小关系的部分的方法有j种

 

 /**
 Create By yzx - stupidboy
 */
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <cmath>
 #include <deque>
 #include <vector>
 #include <queue>
 #include <iostream>
 #include <algorithm>
 #include <map>
 #include <set>
 #include <ctime>
 #include <iomanip>
 using namespace std;
 typedef long long LL;
 typedef double DB;
 #define MIT (2147483647)
 #define INF (1000000001)
 #define MLL (1000000000000000001LL)
 #define sz(x) ((int) (x).size())
 #define clr(x, y) memset(x, y, sizeof(x))
 #define puf push_front
 #define pub push_back
 #define pof pop_front
 #define pob pop_back
 #define ft first
 #define sd second
 #define mk make_pair

 inline int Getint()
 {
     int Ret = ;
     char Ch = ' ';
     bool Flag = ;
     while(!(Ch >= '' && Ch <= ''))
     {
         if(Ch == '-') Flag ^= ;
         Ch = getchar();
     }
     while(Ch >= '' && Ch <= '')
     {
         Ret = Ret *  + Ch - '';
         Ch = getchar();
     }
     return Flag ? -Ret : Ret;
 }

 const int N = ;
 int n;
 int dp[N][N], ans[N];

 inline void Init()
 {
     dp[][] = ;
     for(int i = ; i <= ; i++)
         for(int j = ; j <= i; j++)
             dp[i][j] = dp[i - ][j] * j + dp[i - ][j - ] * j;
     for(int i = ; i <= ; i++)
         for(int j = ; j <= i; j++)
             ans[i] += dp[i][j];
 }

 inline void Solve();

 inline void Input()
 {
     Init();
     while(scanf("%d", &n) == )
     {
         if(n == -) return;
         Solve();
     }
 }

 inline void Solve()
 {
     printf("%d\n", ans[n]);
 }

 int main()
 {
     freopen("a.in", "r", stdin);
     Input();
     //Solve();
     return ;
 }