Max History CodeForces - 938E (组合计数)

You are given an array a of length n. We define f_a the following way:

• Initially f_a = 0, M = 1;
• for every 2 ≤ i ≤ n if a_M < a_i then we set f_a = f_a + a_M and then set M = i.

Calculate the sum of f_a over all n! permutations of the array a modulo 10⁹ + 7.

Note: two elements are considered different if their indices differ, so for every array a there are exactly n! permutations.

先排好序, 考虑$a[i]$的贡献, 显然除了$a[i]$以外所有$\ge a[i]$的数要全部排在$a[i]$前面才能有贡献, 即$\binom{n}{n-i+1}(n-i)!(i-1)!=\frac{n!}{n-i+1}$

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head

#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif

int n, a[N], f[N];

int main() {
	scanf("%d", &n);
	ll fac = 1, ans = 0;
	REP(i,1,n) scanf("%d",a+i), fac=fac*i%P;
	sort(a+1,a+1+n);
	REP(i,1,n) {
		if (a[i]==a[n]) break;
		if (a[i]==a[i-1]) ans+=f[i]=f[i-1];
		else ans+=f[i]=fac*inv(n-i+1)%P*a[i]%P;
	}
	printf("%lld\n",ans%P);
}