牛客网多校训练第一场 F

链接：

https://www.nowcoder.com/acm/contest/139/F

题意：

分析：

转载自：http://tokitsukaze.live/2018/07/19/2018niuke1.F/

代码：

#include <cstdio>
#include <cassert>
#include <algorithm>
using namespace std;

/// 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-1);
    }
    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;
    }
}

typedef long long int LLI;
const LLI MOD = polysum::mod;
const int UP = 1e3 + 5;
LLI a[UP], b[UP];

int main() {
    polysum::init(UP);
    int n;
    while(~scanf("%d", &n)) {
        for(int i = 1; i <= n; i++) scanf("%lld", &a[i]);
        sort(a+1, a+n+1);
        LLI ans = 0, prod = 1;
        for(int i = 1; i <= n; i++) {
            if(a[i] == a[i-1]) {
                prod = prod * a[i] % MOD;
                continue;
            }
            for(int x = 1; x <= n-i+1; x++) {
                b[x] = (polysum::powmod(x, n-i+1) - polysum::powmod(x-1, n-i+1) + MOD) % MOD * x % MOD;
            }
            LLI temp = (polysum::polysum(a[i]+1, b, n-i+1) - polysum::polysum(a[i-1]+1, b, n-i+1) + MOD) % MOD;
            ans = (ans + prod * temp % MOD) % MOD;
            prod = prod * a[i] % MOD;
        }
        printf("%lld
", ans);
    }
    return 0;
}

拉格朗日插值法模板（杜教版）：

/// 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-1);
    }
    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;
    }
}