hdu5322 Hope(dp)

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Hope

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 116    Accepted Submission(s): 44

Problem Description

Hope is a good thing, which can help you conquer obstacles in your life, just keep fighting, and solve the problem below.

In mathematics, the notion of permutation relates to the act of arranging all the members of a set into some sequence or order, or if the set is already ordered, rearranging (reordering) its elements, a process called permuting. These differ from combinations, which are selections of some members of a set where order is disregarded. For example, written as tuples, there are six permutations of the set {1,2,3}, namely: (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), and (3,2,1). These are all the possible orderings of this three element set. As another example, an anagram of a word, all of whose letters are different, is a permutation of its letters. In this example, the letters are already ordered in the original word and the anagram is a reordering of the letters. 
There is a permutation A1,A2,...An, now we define its value as below:
For each Ai, if there exists a minimum j satisfies j>i and Aj>Ai , then connect an edge between Ai and Aj , so after we connect all the edges, there is a graph G, calculate the product of the number of nodes in each component as an integer P. The permutation value is P * P.Now, Mr. Zstu wants to know the sum of all the permutation value of n. In case the answer is very big, please output the answer mod 998244353.
Just in case some of you can’t understand, all the permutations of 3 are
1 2 3
1 3 2
2 3 1
2 1 3
3 1 2
3 2 1

 

Input

There are multiple test cases.
There are no more than 10000 test cases.
Each test case is an integer n(1≤n≤100000).

 

Output

For each test case, output the answer as described above.

 

Sample Input

1
2

 

Sample Output

1
5

dp

 #include <bits/stdc++.h>
 using namespace std;
 #define MOD 998244353
 long long dp[];
 int main()
 {
     int n;
     dp[] = ;
     dp[] = ;
     dp[] = ;
     for(long long i=;i<=;i++)dp[i+] = (((*i+)*dp[i+] - (i+)*(*i+)%MOD*dp[i+] + (i+)*(i+)%MOD*i%MOD*dp[i])%MOD + MOD )%MOD;
     while(scanf("%d",&n)!=EOF) {
         printf("%I64d\n", dp[n]);
     }
     return ;
 }