hdu 4472 Count(简单递推)
2017-10-05 12:38
537 查看
Count
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2465 Accepted Submission(s): 1645
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
1
2
3
40
50
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
分析:简单递推
代码如下:
#include<iostream> #include<cstdio> #include<cstring> using namespace std; const int Mod=1e9+7; long long dp[1005]; void init() { dp[1]=1; dp[2]=1; dp[3]=2; for(int i=4;i<=1000;i++) { dp[i]=0; for(int j=1;j<=i;j++) { if((i-1)%j==0) { dp[i]+=dp[j]; dp[i]%=Mod; } } } } int main() { int n; int inde=0; init(); while(~scanf("%d",&n)) { printf("Case %d: %lld\n",++inde,dp ); } return 0; }
相关文章推荐
- hdu 4472 Count 递推
- HDU-4472 Count 递推
- hdu 4472 Count(递推即dp)
- HDU 4472 count(递推)
- HDU 4472 Count 递推
- hdu 4472 Count (递推)
- hdu 4472 Count (递推)
- HDU 4472 Count(数学 递归)
- HDU 2501 Tiling_easy version(简单递推)
- HDU 4472 Count (DP)
- Hdu 4472 Count 【动态规划】
- HDU 4472 Count
- [递推简单dp]-hdu 2084 数塔
- [dp] hdu 4472 Count
- HDU-超级阶梯(简单递推)
- HDOJ 4472 Count(递推)
- hdu 2046 简单递推
- [递推简单dp]-hdu 2041 超级楼梯
- HDU 2563 统计问题【简单递推】
- HDU 3485 Count 101(简单DP)