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最新推荐文章于 2017-10-04 14:10:53 发布

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最新推荐文章于 2017-10-04 14:10:53 发布

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分类专栏： hdu dp 文章标签：树状dp

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Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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Description

Prof. Tigris is the head of an archaeological team who is currently in charge of an excavation in a site of ancient relics.
This site contains relics of a village where civilization once flourished. One night, examining a writing record, you find some text meaningful to you. It reads as follows.
“Our village is of glory and harmony. Our relationships are constructed in such a way that everyone except the village headman has exactly one direct boss and nobody will be the boss of himself, the boss of boss of himself, etc. Everyone expect the headman is considered as his boss’s subordinate. We call it relationship configuration. The village headman is at level 0, his subordinates are at level 1, and his subordinates’ subordinates are at level 2, etc. Our relationship configuration is harmonious because all people at same level have the same number of subordinates. Therefore our relationship is …”
The record ends here. Prof. Tigris now wonder how many different harmonious relationship configurations can exist. He only cares about the holistic shape of configuration, so two configurations are considered identical if and only if there’s a bijection of n people that transforms one configuration into another one.
Please see the illustrations below for explanation when n = 2 and n = 4.

The result might be very large, so you should take module operation with modules 10 ⁹ +7 before print your answer.

Input

There are several test cases.
For each test case there is a single line containing only one integer n (1 ≤ n ≤ 1000).
Input is terminated by EOF.

Output

For each test case, output one line “Case X: Y” where X is the test case number (starting from 1) and Y is the desired answer.

Sample Input

     
      1
2
3
40
50
600
700
     
     
      很简单的dp题，在这题中，一定要注意，对称性！一定要想想树是可以怎么生成，在这里，我们可以发现，在这个问题中，一个小树，如果乘个j再加一个树根，不就可以长成一个大树了么？我们就可以用这个规律，dp[i]+=dp[(i-1)/j];dp[i]表示有i个结点时的排序数！也就是为(i-1)/j的树*一个任意数j再加一个树根，想想看，是不是这样！
     
     
      #include <iostream>
#include <stdio.h>
using namespace std;
#define  modnum 1000000007
int dp[1005];
int main()
{
   int i,j,tcase,n;
   tcase=1;
   dp[1]=1;
    for(i=2;i<=1000;i++)
   {
       for(j=1;j<i;j++)
       {
           if((i-1)%j==0)
           {
               dp[i]+=dp[(i-1)/j];
               dp[i]%=modnum;
           }
       }
   }
   while(scanf("%d",&n)!=EOF)
   {
       printf("Case %d: %d\n",tcase++,dp[n]);
   }

    return 0;
}