竞赛讨论区 > 牛客网暑期ACM多校训练营(第一场)F 题解
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tokitsukaze
编辑于 2018-07-22 17:29
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牛客网暑期ACM多校训练营(第一场)F 题解



题解:


 


至于杜教的拉格朗日插值板子,是从这里扒的:http://codeforces.com/contest/995/submission/39615335


代码:
#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define mem(a,b) memset((a),(b),sizeof(a))
#define MP make_pair
#define pb push_back
#define fi first
#define se second
#define sz(x) (int)x.size()
#define all(x) x.begin(),x.end()
#define _GLIBCXX_PERMIT_BACKWARD_HASH
#include <ext/hash_map>
using namespace __gnu_cxx;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll,ll> PLL;
typedef vector<int> VI;
typedef vector<ll> VL;
struct str_hash{size_t operator()(const string& str)const{return __stl_hash_string(str.c_str());}};
const int INF=0x3f3f3f3f;
const ll LLINF=0x3f3f3f3f3f3f3f3f;
const double PI=acos(-1.0);
const double eps=1e-4;
const int MAX=2e5+10;
const ll mod=1e9+7;
/****************************************  head  ****************************************/
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--)
    const int D=2010;
    ll a[D],f[D],g[D],p[D],p1[D],p2[D],b[D],h[D][2],C[D];
    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;}
    ll calcn(int d,ll *a,ll n) { // a[0].. a[d]  a[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) {
        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 m,ll *a,ll n) { // a[0].. a[m] \sum_{i=0}^{n-1} a[i]
        ll b[D];
        for(int i=0;i<=m;i++) b[i]=a[i];
        b[m+1]=calcn(m,b,m+1);
        rep(i,1,m+2) b[i]=(b[i-1]+b[i])%mod;
        return calcn(m+1,b,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;
    }
} // polysum::init();
ll pow2(ll a,ll b)
{
    ll res=1;
    while(b)
    {
        if(b&1) res=res*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    return res;
}
int main()
{
    int t,n,i,j;
    ll ans,a[1111],now,b[1111];
    polysum::init(1010);
    while(~scanf("%d",&n))
    {
        for(i=1;i<=n;i++) scanf("%lld",&a[i]);
        ans=0;
        sort(a+1,a+1+n);
        a[0]=0;
        now=1;
        for(i=1;i<=n;i++)
        {
            if(a[i]==a[i-1])
            {
                (now*=a[i])%=mod;
                continue;
            }
            b[0]=0;
            for(j=1;j<=n-i+1;j++)
            {
                b[j]=j*((pow2(j,n-i+1)-pow2(j-1,n-i+1)%mod)+mod)%mod;
            }
            ll tmp=((polysum::polysum(n-i+1,b,a[i]+1)-polysum::polysum(n-i+1,b,a[i-1]+1))%mod+mod)%mod;
            (ans+=tmp*now%mod)%=mod;
            (now*=a[i])%=mod;
        }
        printf("%lld\n",ans);
    }
    return 0;
}




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