Dream_Chaser队训练赛第一场 I题

题目来自2012成都区域赛

I - Count

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

Submit Status Practice HDU 4472

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

600

700

Sample Output

Case 1: 1

Case 2: 1

Case 3: 2

Case 4: 924

Case 5: 1998

Case 6: 315478277

Case 7: 825219749

思路：递推式类似fib数列，加记忆化或者递推即可。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<vector>
 7 #include<stack>
 8 #include<queue>
 9 #include<map>
10 #include<set>
11 #include<math.h>
12 #include<string>
13 #include<cctype>
14
15 using namespace std;
16
17 const int maxn=1000100;
18 const int INF=(1<<29);
19 const double EPS=0.00000001;
20 typedef long long ll;
21 const ll p=1000000000+7;
22
23 ll n;
24 ll F[maxn];
25
26 ll f(ll n)
27 {
28     if(F[n]!=-1) return F[n];
29     if(n==1) return F[n]=1;
30     if(n==2) return F[n]=1;
31     if(n==3) return F[n]=2;
32     //return f(n-1)+(n%2==1?1:0)+1;
33     ll res=0;
34     res=(res%p+f(n-1)%p+1)%p;
35     for(int i=2;i<=n/2;i++){
36         if(n%i==1) res=(res%p+f(n/i)%p)%p;
37     }
38     return F[n]=res%p;
39 }
40
41 void play()
42 {
43     for(int i=1;i<=1010;i++){
44         printf("%I64d,",f(i));
45         if(i%7==0) printf("\n");
46     }
47 }
48
49 int main()
50 {
51     memset(F,-1,sizeof(F));
52     int tag=1;
53     while(cin>>n){
54         printf("Case %d: %I64d\n",tag++,f(n));
55     }
56     return 0;
57 }

View Code

转载于:https://www.cnblogs.com/--560/p/4543047.html