bzoj3028(生成函数)

题目链接

https://www.lydsy.com/JudgeOnline/problem.php?id=3028

题解

沙雕生成函数题,对每个物品构造生成函数相乘,得

疯狂化简之后得到

利用广义二项式展开有

带入上式得答案为 $\binom{k+2}{3}$




代码

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/**
*         ┏┓    ┏┓
*         ┏┛┗━━━━━━━┛┗━━━┓
*         ┃       ┃  
*         ┃   ━    ┃
*         ┃ >   < ┃
*         ┃       ┃
*         ┃... ⌒ ...  ┃
*         ┃ ┃
*         ┗━┓ ┏━┛
*          ┃ ┃ Code is far away from bug with the animal protecting          
*          ┃ ┃ 神兽保佑,代码无bug
*          ┃ ┃           
*          ┃ ┃       
*          ┃ ┃
*          ┃ ┃           
*          ┃ ┗━━━┓
*          ┃ ┣┓
*          ┃ ┏┛
*          ┗┓┓┏━━━━━━━━┳┓┏┛
*           ┃┫┫ ┃┫┫
*           ┗┻┛ ┗┻┛
*/

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<stack>
#include<cmath>
#include<set>
#include<bitset>
#include<complex>
#include<cstdlib>
#include<assert.h>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define eps 1e-8
#define succ(x) (1<<x)
#define lowbit(x) (x&(-x))
#define mid (x+y>>1)
#define sqr(x) ((x)*(x))
#define NM 8005
#define nm 400005
const double pi=acos(-1);
const int inf=10007;
using namespace std;
ll read(){
ll x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return f*x;
}





ll t;
int n;
char _s[NM];

int main(){
scanf("%s",_s+1);
n=strlen(_s+1);
inc(i,1,n)t=(t*10+_s[i]-'0')%inf;
printf("%lld\n",t*(t+1)*(t+2)/6%inf);
}