ICPC2019南昌邀請賽 B. Polynomial

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題解

這題我做麻煩了，我是直接把係數$a_0,a_1,\dots,a_n$求出來，然後再得到$f(n+1)$，然後再利用拉格朗日插值求$F(M) = \sum_{i=0}^M f(i)$

但是其實可以直接利用已知的$f(0),f(1),\dots,f(n)$求出$f(n+1)$。

不管怎樣，最後求出$F(x)$在$0,1,\dots,n+1$處的值之後，就可以對每次詢問直接插值求$F(R)$和$F(L-1)$了

代碼

#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(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#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;
}
#define mod 9999991ll
struct LagrangeInterpolation
{
    ll _fact[maxn], inv[maxn], pref[maxn], suf[maxn];
    ll calc(ll n, ll *y, ll X)
    {
        ll i, ans=0;
        X%=mod;
        inv[1]=1; rep(i,2,n)inv[i]=inv[mod%i]*(mod-mod/i)%mod;
        _fact[0]=1; rep(i,1,n)_fact[i]=_fact[i-1]*inv[i]%mod;
        pref[0]=X, suf[n+1]=1;
        rep(i,1,n)pref[i]=pref[i-1]*(X-i)%mod;
        drep(i,n,0)suf[i]=suf[i+1]*(X-i)%mod;
        rep(i,0,n)
        {
            ll t=suf[i+1];
            if(i)(t*=pref[i-1])%=mod;
            (t*=_fact[i]*_fact[n-i]%mod)%=mod;
            if(n-i&1)t=-t;
            (ans+=t*y[i]%mod)%=mod;
        }
        return ans;
    }
}lagrange;
ll y[maxn], n, m, L, R, a[maxn], f[maxn], C[maxn], P[maxn], inv[maxn];
int main()
{
    ll T=read();
    while(T--)
    {
        ll i, j;
        n=read(), m=read();
        inv[1]=1; rep(i,2,n)inv[i]=inv[mod%i]*(mod-mod/i)%mod;
        rep(i,0,n)f[i]=read();
        rep(i,0,n)a[i]=0;
        rep(i,0,n+1)P[i]=0;
        P[0]=1;
        rep(i,0,n)
        {
            drep(j,n+1,1)
            {
                P[j] = P[j-1]-i*P[j];
                P[j]%=mod;
            }
            P[0]=-i*P[0]%mod;
        }
        rep(i,0,n)
        {
            ll t = f[i];
            rep(j,0,n)if(i!=j)
            {
                if(j<i)t*=inv[i-j];
                else t*=-inv[j-i];
                t%=mod;
            }
            if(i==0)
            {
                rep(j,0,n)C[j]=P[j+1];
            }
            else
            {
                C[0]=0;
                rep(j,1,n)C[j]=(C[j-1]-P[j])*inv[i]%mod;
            }
            rep(j,0,n)(a[j]+=C[j]*t)%=mod;
        }
        rep(i,0,n+1)
        {
            y[i]=0;
            ll x=1;
            rep(j,0,n)
            {
                (y[i]+=a[j]*x)%=mod;
                (x*=i)%=mod;
            }
            if(i)(y[i]+=y[i-1])%=mod;
        }
        while(m--)
        {
            L=read(), R=read();
            ll ans = lagrange.calc(n+1,y,R) - lagrange.calc(n+1,y,L-1);
            printf("%lld\n",(ans%mod+mod)%mod);
        }
    }
    return 0;
}