JZOJ 6977 去南极

又是 dirty works……（

首先有 DP fn=sn+1(n3)n∑i=1(i−1)(n−i)(fk+fn−k)

随便化一下可得 (n3)fn=(n3)sn+2n∑i=1(n−i−1)ifi

写成生成函数的形式就是 F(3)(x)=S(3)(x)+12(1−x)2F′(x)(1−x)3F(3)−12(1−x)F′=S(3)

我也不知道为什么定义算子 θF(x)=(1−x)F′(x)

（注意这里的定义和题解不一样）

然后可以发现 θ2F=(1−x)2F″−(1−x)F′θ3F=(1−x)3F(3)−3(1−x)2F″+(1−x)F′

因此 (θ3+3θ2−10θ)F=S(3)θ(θ−2)(θ+5)F=(1−x)3S(3)

令 Q=(θ+5)F,P=(θ−2)Q，则有 {θP=(1−x)3S(3)(θ−2)Q=P(θ+5)F=Q

提取系数得 {(n+1)pn+1=npn+▽3sn+3(n+3)3_(n+1)qn+1=(n+2)qn+pn(n+1)fn=(n−5)fn+qn

从第一条式子有 npn=▽2sn+2(n+2)3_

再代入第二条式子，展开得 qn+1n+2=qm+1m+2+n+1∑i=m+3▽2si

再代入第三条式子，展开得 (n6)fn=(m+16)fm+1+qm+1m+2n∑i=m+2(i6)+n∑i=m+2(i6)i∑j=m+3▽2sj

显然有 fm+1=sm+1,fm+2=sm+2

然后 pm+1=(m+2)fm+2−(m−4)fm+1

再一通乱算可以得到 fn=sn+12(n6)n−1∑i=m+1fi(i+1)(i+2)((n+17)−(i+27))

使用线段树维护即可。

代码：

#include <cstdio>
#include <utility>
#include <algorithm>
using namespace std;
const int N = 2e5 + 1;
const int mod = 998244353;
int L,q;
int a[N + 5],inv[N + 5];
int s1[N + 5],s2[N + 5];
namespace SEG
{
    #define ls (p << 1)
    #define rs (ls | 1)
    struct node
    {
        int sum1,sum2;
        int tag;
    } seg[(N << 2) + 5];
    inline void push(int p,int tl,int tr)
    {
        int mid = tl + tr >> 1;
        if(seg[p].tag)
        {
            seg[ls].sum1 = (seg[ls].sum1 + (long long)seg[p].tag * (s1[mid] - s1[tl - 1] + mod)) % mod,
            seg[ls].sum2 = (seg[ls].sum2 + (long long)seg[p].tag * (s2[mid] - s2[tl - 1] + mod)) % mod,
            seg[ls].tag = (seg[ls].tag + seg[p].tag) % mod;
            seg[rs].sum1 = (seg[rs].sum1 + (long long)seg[p].tag * (s1[tr] - s1[mid] + mod)) % mod,
            seg[rs].sum2 = (seg[rs].sum2 + (long long)seg[p].tag * (s2[tr] - s2[mid] + mod)) % mod,
            seg[rs].tag = (seg[rs].tag + seg[p].tag) % mod;
            seg[p].tag = 0;
        }
    }
    void update(int l,int r,int k,int p,int tl,int tr)
    {
        if(l <= tl && tr <= r)
        {
            seg[p].sum1 = (seg[p].sum1 + (long long)k * (s1[tr] - s1[tl - 1] + mod)) % mod,
            seg[p].sum2 = (seg[p].sum2 + (long long)k * (s2[tr] - s2[tl - 1] + mod)) % mod,
            seg[p].tag = (seg[p].tag + k) % mod;
            return ;
        }
        push(p,tl,tr);
        int mid = tl + tr >> 1;
        l <= mid && (update(l,r,k,ls,tl,mid),1);
        r > mid && (update(l,r,k,rs,mid + 1,tr),1);
        seg[p].sum1 = (seg[ls].sum1 + seg[rs].sum1) % mod,
        seg[p].sum2 = (seg[ls].sum2 + seg[rs].sum2) % mod;
    }
    inline pair<int,int> operator+(const pair<int,int> &a,const pair<int,int> &b)
    {
        return make_pair((a.first + b.first) % mod,(a.second + b.second) % mod);
    }
    pair<int,int> query(int l,int r,int p,int tl,int tr)
    {
        if(l <= tl && tr <= r)
            return make_pair(seg[p].sum1,seg[p].sum2);
        push(p,tl,tr);
        int mid = tl + tr >> 1;
        pair<int,int> ret(0,0);
        l <= mid && (ret = ret + query(l,r,ls,tl,mid),1);
        r > mid && (ret = ret + query(l,r,rs,mid + 1,tr),1);
        return ret;
    }
}
namespace BIT
{
    #define lowbit(x) ((x) & -(x))
    int c[N + 5];
    inline void update(int x,int k)
    {
        for(;x <= L;x += lowbit(x))
            c[x] = (c[x] + k) % mod;
    }
    inline void update(int l,int r,int k)
    {
        update(l,k),update(r + 1,(mod - k) % mod);
    }
    inline int query(int x)
    {
        int ret = 0;
        for(;x;x -= lowbit(x))
            ret = (ret + c[x]) % mod;
        return ret;
    }
}
int ans;
int main()
{
    freopen("space.in","r",stdin),freopen("space.out","w",stdout);
    scanf("%d%d",&L,&q),inv[1] = 1;
    for(register int i = 2;i <= N;++i)
        inv[i] = (long long)(mod - mod / i) * inv[mod % i] % mod;
    for(register int i = 1;i <= L;++i)
        scanf("%d",a + i),
        s1[i] = (s1[i - 1] + (long long)inv[i + 1] * inv[i + 2]) % mod,
        s2[i] = (s2[i - 1] + (long long)i * (i - 1) % mod * (i - 2) % mod * (i - 3) % mod * (i - 4) % mod * inv[5040]) % mod;
    for(register int i = 1;i <= L;++i)
        SEG::update(i,i,a[i],1,1,L),
        BIT::update(i,i,a[i]);
    for(int op,l,r,k,n,m;q;--q)
    {
        scanf("%d",&op);
        if(!op)
            scanf("%d%d%d",&l,&r,&k),
            SEG::update(l,r,k,1,1,L),
            BIT::update(l,r,k);
        else
        {
            scanf("%d%d",&n,&m);
            if(n <= max(m,5))
                printf("%d\n",BIT::query(n));
            else
            {
                pair<int,int> temp = SEG::query(m + 1,n - 1,1,1,L);
                ans = (long long)temp.first * (n + 1) % mod * n % mod * (n - 1) % mod * (n - 2) % mod * (n - 3) % mod * (n - 4) % mod * (n - 5) % mod * inv[5040] % mod,
                ans = (ans - temp.second + mod) % mod,
                ans = (12LL * 720 * ans % mod * inv[n] % mod * inv[n - 1] % mod * inv[n - 2] % mod * inv[n - 3] % mod * inv[n - 4] % mod * inv[n - 5] + BIT::query(n)) % mod;
                printf("%d\n",ans);
            }
        }
    }
}