hdu 5322 Hope

Hope

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 456 Accepted Submission(s): 251

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

Author
ZSTU

Source
2015 Multi-University Training Contest 3

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【分析】

題意：

給定n，考慮一個1,2,…,n的排列A[1],A[2],…,A[n]，對於每個i，選取最小的j(若存在)使得j>i且A[j]>A[i]，則在i到j之間連一條邊，記P爲圖中所有連通塊的大小之積，定義P*P爲這個排列的permutation value，求出所有1,2,…,n的排列的permutation value之和對998244353取模的值。

[題意 copy from quailty ]

記n 的答案爲dp[n]

考慮枚舉數字n 放在第幾個位置

dp[n]=∑ni=1(i−1)!C(n−1,i−1)∗i2∗dp[n−i]

考慮展開C，得到：

dp[n]=(n−1)!∑n−1i=0(n−i)2i!

這是個卷積…CDQ+NTT…
初始化 dp[0]=1

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【代碼】

//hdu 5322 Hope
#include<bits/stdc++.h>
#define M 100000
#define ll long long
#define fo(i,j,k) for(int i=j;i<=k;i++)
using namespace std;
const int mod=998244353;
const int mxn=400005;
int n,m,T,L;
int a[mxn],b[mxn],dp[mxn];
int fac[mxn],inv[mxn],R[mxn];
inline int read()
{
    int x=0;char ch=getchar();
    while(ch<'0' || ch>'9') ch=getchar();
    while(ch>='0' && ch<='9') x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
    return x;
}
inline int power(int x,int k)
{
    int res=1;
    while(k)
    {
        if(k&1) res=(ll)res*x%mod;
        x=(ll)x*x%mod,k>>=1;
    }
    return res;
}
inline void NTT(int *a,int f)
{
    fo(i,0,n-1) if(i<R[i]) swap(a[i],a[R[i]]);
    for(int i=1;i<n;i<<=1)
    {
        int wn=power(3,(mod-1)/(i<<1));
        for(int j=0;j<n;j+=(i<<1))
        {
            int w=1;
            for(int k=0;k<i;k++,w=(ll)w*wn%mod)
            {
                int x=a[j+k],y=(ll)w*a[j+k+i]%mod;
                a[j+k]=(x+y)%mod;
                a[j+k+i]=(x-y+mod)%mod;
            }
        }
    }
    if(f==-1)
    {
        reverse(a+1,a+n);
        int rev=power(n,mod-2);
        fo(i,0,n-1) a[i]=(ll)a[i]*rev%mod;
    }
}
inline void CDQ(int l,int r)
{
    if(l==r)
    {
        if(l==0) return;
        dp[l]=(ll)dp[l]*fac[l-1]%mod;
        return;
    }
    int mid=l+r>>1;
    CDQ(l,mid);

    n=r-l,m=n+n;
    fo(i,0,4*n) a[i]=b[i]=0;
    fo(i,l,mid) a[i-l]=(ll)dp[i]*inv[i]%mod;
    fo(i,0,r-l) b[i]=(ll)i*i%mod;
    for(n=1,L=0;n<=m;n<<=1) L++;
    fo(i,0,n-1) R[i]=(R[i>>1]>>1)|((i&1)<<L-1);
    NTT(a,1),NTT(b,1);
    fo(i,0,n) a[i]=(ll)a[i]*b[i]%mod;
    NTT(a,-1);
    fo(i,mid+1,r) dp[i]=(dp[i]+a[i-l])%mod;

    CDQ(mid+1,r);
}
inline void init()
{
    fac[0]=inv[0]=inv[1]=1;
    fo(i,1,M) fac[i]=(ll)fac[i-1]*i%mod;
    fo(i,2,M) inv[i]=(ll)(mod-mod/i)*inv[mod%i]%mod;
    fo(i,1,M) inv[i]=(ll)inv[i]*inv[i-1]%mod;
    dp[0]=1,CDQ(0,M);
}
int main()
{
    init();
    while(scanf("%d",&n)!=EOF)
      printf("%d\n",dp[n]);
    return 0;
}