杜教的板子

拉格朗日插值:

/// 注意mod，使用前须调用一次 polysum::init(int M);
namespace polysum {
    #define rep(i,a,n) for (int i=a;i<n;i++)
    #define per(i,a,n) for (int i=n-1;i>=a;i--)
    typedef long long ll;
    const ll mod=1e9+7; /// 取模值
    ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}

    const int D=101000; /// 最高次限制
    ll a[D],f[D],g[D],p[D],p1[D],p2[D],b[D],h[D][2],C[D];
    ll calcn(int d,ll *a,ll n) {
        if (n<=d) return a[n];
        p1[0]=p2[0]=1;
        rep(i,0,d+1) {
            ll t=(n-i+mod)%mod;
            p1[i+1]=p1[i]*t%mod;
        }
        rep(i,0,d+1) {
            ll t=(n-d+i+mod)%mod;
            p2[i+1]=p2[i]*t%mod;
        }
        ll ans=0;
        rep(i,0,d+1) {
            ll t=g[i]*g[d-i]%mod*p1[i]%mod*p2[d-i]%mod*a[i]%mod;
            if ((d-i)&1) ans=(ans-t+mod)%mod;
            else ans=(ans+t)%mod;
        }
        return ans;
    }
    void init(int M) { /// M：最高次
        f[0]=f[1]=g[0]=g[1]=1;
        rep(i,2,M+5) f[i]=f[i-1]*i%mod;
        g[M+4]=powmod(f[M+4],mod-2);
        per(i,1,M+4) g[i]=g[i+1]*(i+1)%mod;
    }
    ll polysum(ll n,ll *arr,ll m) { // a[0].. a[m] \sum_{i=0}^{n} a[i]
        for(int i = 0; i <= m; i++) a[i] = arr[i];
        a[m+1]=calcn(m,a,m+1);
        rep(i,1,m+2) a[i]=(a[i-1]+a[i])%mod;
        return calcn(m+1,a,n);
    }
    ll qpolysum(ll R,ll n,ll *a,ll m) { // a[0].. a[m] \sum_{i=0}^{n-1} a[i]*R^i
        if (R==1) return polysum(n,a,m);
        a[m+1]=calcn(m,a,m+1);
        ll r=powmod(R,mod-2),p3=0,p4=0,c,ans;
        h[0][0]=0;h[0][1]=1;
        rep(i,1,m+2) {
            h[i][0]=(h[i-1][0]+a[i-1])*r%mod;
            h[i][1]=h[i-1][1]*r%mod;
        }
        rep(i,0,m+2) {
            ll t=g[i]*g[m+1-i]%mod;
            if (i&1) p3=((p3-h[i][0]*t)%mod+mod)%mod,p4=((p4-h[i][1]*t)%mod+mod)%mod;
            else p3=(p3+h[i][0]*t)%mod,p4=(p4+h[i][1]*t)%mod;
        }
        c=powmod(p4,mod-2)*(mod-p3)%mod;
        rep(i,0,m+2) h[i][0]=(h[i][0]+h[i][1]*c)%mod;
        rep(i,0,m+2) C[i]=h[i][0];
        ans=(calcn(m,C,n)*powmod(R,n)-c)%mod;
        if (ans<0) ans+=mod;
        return ans;
    }
}

例题:https://ac.nowcoder.com/acm/contest/139/F?&headNav=www

代码:

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define y1 y11
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
//#define mp make_pair
#define pb push_back
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define pdi pair<double, int>
#define pdd pair<double, double>
#define mem(a, b) memset(a, b, sizeof(a))
#define debug(x) cerr << #x << " = " << x << "\n";
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//head


const int N = 1e3 + 5;
const int MOD = 1e9 + 7;
int a[N], n;
LL b[N];
/// 注意mod，使用前须调用一次 polysum::init(int M);
namespace polysum {
    #define rep(i,a,n) for (int i=a;i<n;i++)
    #define per(i,a,n) for (int i=n-1;i>=a;i--)
    typedef long long ll;
    const ll mod=1e9+7; /// 取模值
    ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}

    const int D=101000; /// 最高次限制
    ll a[D],f[D],g[D],p[D],p1[D],p2[D],b[D],h[D][2],C[D];
    ll calcn(int d,ll *a,ll n) {
        if (n<=d) return a[n];
        p1[0]=p2[0]=1;
        rep(i,0,d+1) {
            ll t=(n-i+mod)%mod;
            p1[i+1]=p1[i]*t%mod;
        }
        rep(i,0,d+1) {
            ll t=(n-d+i+mod)%mod;
            p2[i+1]=p2[i]*t%mod;
        }
        ll ans=0;
        rep(i,0,d+1) {
            ll t=g[i]*g[d-i]%mod*p1[i]%mod*p2[d-i]%mod*a[i]%mod;
            if ((d-i)&1) ans=(ans-t+mod)%mod;
            else ans=(ans+t)%mod;
        }
        return ans;
    }
    void init(int M) { /// M：最高次
        f[0]=f[1]=g[0]=g[1]=1;
        rep(i,2,M+5) f[i]=f[i-1]*i%mod;
        g[M+4]=powmod(f[M+4],mod-2);
        per(i,1,M+4) g[i]=g[i+1]*(i+1)%mod;
    }
    ll polysum(ll n,ll *arr,ll m) { // a[0].. a[m] \sum_{i=0}^{n-1} a[i]
        for(int i = 0; i <= m; i++) a[i] = arr[i];
        a[m+1]=calcn(m,a,m+1);
        rep(i,1,m+2) a[i]=(a[i-1]+a[i])%mod;
        return calcn(m+1,a,n);
    }
    ll qpolysum(ll R,ll n,ll *a,ll m) { // a[0].. a[m] \sum_{i=0}^{n-1} a[i]*R^i
        if (R==1) return polysum(n,a,m);
        a[m+1]=calcn(m,a,m+1);
        ll r=powmod(R,mod-2),p3=0,p4=0,c,ans;
        h[0][0]=0;h[0][1]=1;
        rep(i,1,m+2) {
            h[i][0]=(h[i-1][0]+a[i-1])*r%mod;
            h[i][1]=h[i-1][1]*r%mod;
        }
        rep(i,0,m+2) {
            ll t=g[i]*g[m+1-i]%mod;
            if (i&1) p3=((p3-h[i][0]*t)%mod+mod)%mod,p4=((p4-h[i][1]*t)%mod+mod)%mod;
            else p3=(p3+h[i][0]*t)%mod,p4=(p4+h[i][1]*t)%mod;
        }
        c=powmod(p4,mod-2)*(mod-p3)%mod;
        rep(i,0,m+2) h[i][0]=(h[i][0]+h[i][1]*c)%mod;
        rep(i,0,m+2) C[i]=h[i][0];
        ans=(calcn(m,C,n)*powmod(R,n)-c)%mod;
        if (ans<0) ans+=mod;
        return ans;
    }
}
int main() {
    polysum::init(N);
    while(~scanf("%d", &n)) {
        for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
        sort(a+1, a+1+n);
        LL pre = 1, ans = 0;
        for (int i = 1; i <= n; ++i) {
            b[0] = 0;
            for (int j = 1; j <= n-i+1; ++j) b[j] = (j*(polysum::powmod(j, n-i+1) - polysum::powmod(j-1, n-i+1))) % MOD;
            ans = (ans + pre*(polysum::polysum(a[i], b, n-i+1) - polysum::polysum(a[i-1], b, n-i+1))%MOD )%MOD;
            pre = (pre * a[i]) % MOD;
        }
        printf("%lld\n", (ans + MOD) % MOD);
    }
    return 0;
}

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