HDU4472 Count DP

Count

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2148    Accepted Submission(s): 1421

Problem 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 10⁹ +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

1234050600700

 

Sample Output

Case 1: 1Case 2: 1Case 3: 2Case 4: 924Case 5: 1998Case 6: 315478277Case 7: 825219749

 

题意：有n个人，图表示上下级关系，对于同一级的人来说，每个人拥有属下的数量是一样的，问不同的组合方式有多少种。

dp[i]表示当某个状态的最后一层有i个节点的时候，组合的方法数，总共的方法数就是1~n-1种结尾方式组合数的和。

#include <iostream>#include <stdio.h>#include <string>#include <cstring>const __int64 mod=1e9+7;using namespace std; __int64 dp[1009]; int main(){    int n;    int ca=1;    while(~scanf("%d",&n))    {        if(n==1)        {            printf("Case %d: %d\n",ca++,1);            continue;        }        memset(dp,0,sizeof dp);         dp[n-1]=1;        //dp[1]=1;        for(int i=n-1;i>=2;i--)        {            for(int j=i/2;j<i;j++)//向下扩展时候，每次拿j个节点下来，使得拿下的j个节点均匀分在(i-j)个节点下方            {                if(j%(i-j)==0)                {                    dp[(j/(i-j))]=(dp[(j/(i-j))]+dp[i])%mod;                }            }        }        __int64 ans=0;        for(int i=1;i<=n;i++)        {            ans=(ans+dp[i])%mod;             //cout<<dp[i]<<endl;        }       // cout<<ans<<endl;        printf("Case %d: %I64d\n",ca++,ans);    }    return 0;}