题意：

在这里插入图片描述

思考

考虑将每个条件转换成生成函数：
$(1)f1(x)=1+x2+...=11−x2(1)f_1(x)=1+x^2+...=\frac{1}{1-x^2}$
$(2)f2(x)=1+x=1−x21−x(2)f_2(x)=1+x=\frac{1-x^2}{1-x}$
$(3)f3(x)=1+x+x2=1−x31−x(3)f_3(x)=1+x+x^2=\frac{1-x^3}{1-x}$
$(4)f4(x)=x+x3+...=x1−x2(4)f_4(x)=x+x^3+...=\frac{x}{1-x^2}$
$(5)f5(x)=1+x4+...=11−x4(5)f_5(x)=1+x^4+...=\frac{1}{1-x^4}$
$(6)f6(x)=1+x+x2+x3=1−x41−x(6)f_6(x)=1+x+x^2+x^3=\frac{1-x^4}{1-x}$
$(7)f7(x)=1+x=1−x21−x(7)f_7(x)=1+x=\frac{1-x^2}{1-x}$
$(8)f8(x)=1+x3+x6+...=11−x3(8)f_8(x)=1+x^3+x^6+...=\frac{1}{1-x^3}$
将其乘起来，得： $(1−x2)(1−x3)x(1−x4)(1−x2)(1−x2)(1−x)(1−x)(1−x2)(1−x4)(1−x)(1−x)(1−x3)\frac{(1-x^2)(1-x^3)x(1-x^4)(1-x^2)}{(1-x^2)(1-x)(1-x)(1-x^2)(1-x^4)(1-x)(1-x)(1-x^3)}$
约分可得 $x(1−x)4\frac{x}{(1-x)^4}$ 。
由广义二项式定理 $1(1−x)m+1=∑n≥0(n+mn)xn\frac{1}{(1-x)^{m+1}}=\sum_{n\ge 0}\binom{n+m}{n}x^n$ 得 $x(1−x)4=∑n≥0(n+3n)xn+1=∑n≥1(n+2n−1)xn\frac{x}{(1-x)^4}=\sum_{n\ge0}\binom{n+3}{n}x^{n+1}=\sum_{n\ge1}\binom{n+2}{n-1}x^n$ ，所以第 $n$ 项的答案为 $(n+2n−1)=(n+23)=(n+2)(n+1)(n)6\binom{n+2}{n-1}=\binom{n+2}{3}=\frac{(n+2)(n+1)(n)}{6}$ ，取个逆元直接计算即可。

//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#include<cassert>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid ((tr[u].l+tr[u].r)>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1000010,mod=10007,INF=0x3f3f3f3f;
const double eps=1e-6;string s;LL qmi(LL a,LL b) {LL ans=1;while(b) {if(b&1) ans=ans*a%mod;a=a*a%mod;b>>=1;}return ans%mod;
}int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);cin>>s;int sum=0;for(int i=0;i<s.length();i++) sum=sum*10+s[i]-'0',sum%=mod;sum+=2; sum%=mod;int a=sum,b=(sum-1+mod)%mod,c=(sum-2+mod)%mod;printf("%lld\n",1ll*a*b*c%mod*qmi(6,mod-2)%mod);return 0;
}
/**/