nowcoder17694 Rikka with Prefix Sum

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Well this question

Observing a $1$ After several prefixes and impact on the back of:

$\begin{matrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 \\ 1 & 3 & 6 & 10\\ \end{matrix}$

In fact, the law has come out, a $1$ on the back distance $d$ positions after $cnt$ times prefixes and contributions $C_ {cnt + d-1} ^ {d}$, Sum and difference are reciprocal process, interval modification is the first difference at a single point and then modify the query interval is the sum once and then a single point of inquiry.

Violence count just fine

Code

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(_,__) for(_=1;_<=(__);_++)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
struct EasyMath
{
    ll prime[maxn], phi[maxn], mu[maxn];
    bool mark[maxn];
    ll fastpow(ll a, ll b, ll c)
    {
        ll t(a%c), ans(1ll);
        for(;b;b>>=1,t=t*t%c)if(b&1)ans=ans*t%c;
        return ans;
    }
    void shai(ll N)
    {
        ll i, j;
        for(i=2;i<=N;i++)mark[i]=false;
        *prime=0;
        phi[1]=mu[1]=1;
        for(i=2;i<=N;i++)
        {
            if(!mark[i])prime[++*prime]=i, mu[i]=-1, phi[i]=i-1;
            for(j=1;j<=*prime and i*prime[j]<=N;j++)
            {
                mark[i*prime[j]]=true;
                if(i%prime[j]==0)
                {
                    phi[i*prime[j]]=phi[i]*prime[j];
                    break;
                }
                mu[i*prime[j]]=-mu[i];
                phi[i*prime[j]]=phi[i]*(prime[j]-1);
            }
        }
    }
    ll inv(ll x, ll p)  //p是素数
    {return fastpow(x%p,p-2,p);}
}em;
#define mod 998244353ll
ll n, m, fact[maxn], _fact[maxn], type[maxn], L[maxn], R[maxn], w[maxn];
ll C(ll n, ll m)
{
    return fact[n]*_fact[m]%mod*_fact[n-m]%mod;
}
ll calc(ll d, ll cnt)
{
    if(d<0)return 0;
    return C(cnt+d-1,d);
}
int main()
{
    ll i, j, T=read();
    fact[0]=_fact[0]=1;
    rep(i,1e6)fact[i]=fact[i-1]*i%mod, _fact[i]=em.fastpow(fact[i],mod-2,mod);
    while(T--)
    {
        n=read(), m=read();
        rep(i,m)
        {
            type[i]=read();
            if(type[i]!=2)L[i]=read(), R[i]=read();
            if(type[i]==1)w[i]=read();
            if(type[i]==3)
            {
                ll cnt=0, ans=0;
                for(j=i-1;j;j--)
                {
                    if(type[j]==1)
                    {
                        ans += w[j]*( -calc(L[i]-1-L[j],cnt+2) + calc(R[i]-L[j],cnt+2) + calc(L[i]-1-R[j]-1,cnt+2) - calc(R[i]-R[j]-1,cnt+2) );
                        ans %= mod;
                    }
                    if(type[j]==2)cnt++;
                }
                printf("%lld\n",(ans+mod)%mod);
            }
        }
    }
    return 0;
}