hdu4472(DP)

 

Count

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1063    Accepted Submission(s): 695

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

 

Source

2012 Asia Chengdu Regional Contest

 

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        本题要求子树形状相同的树的个数。

      对于拥有n个节点的树。我们可以枚举子树的个数。要是子树的形状相同，必须保证子树的节点数==（n-1）/i；其中i为子树的个数。这样可以用DP的思想求。

     dp[n]=∑dp[i];其中(n-1)能整出i.

#include<iostream>#include<cstdio>using namespace std;int dp[1000+10];const int mod=1000000000+7;int func(int n){    int i;    if(dp[n])        return dp[n];    else if(n==1)    {        return 1;    }    for(i=1;i<=n-1;i++)    {        if((n-1)%i==0)            dp[n]+=func((n-1)/i);        dp[n]=dp[n]%mod;    }    return dp[n];}int main(){    int n,i,tag=0;    int ans;    memset(dp,0,sizeof(dp));    while(~scanf("%d",&n))    {        ans=func(n);        printf("Case %d: %d\n",++tag,ans);    }    return 0;}