Bzoj4710 [Jsoi2011] sub-specialty

Time Limit:10 Sec Memory limit:128 MB
submit:96 solved:62
[Submit] [Status] [Discuss] Descriptionjyy led participated in a number of ACM/ICPC competitions, brought back a lot of native products, to be divided into the laboratory students. Jyy want to know, how many different ways to divide these specialties into n classmates? Of course, jyy do not want any of the students because they did not get a special product and feel lost, so every classmate must at least a specialty. For example, Jyy brought 2 bags of twist and 1 bags of steamed buns to a and B two students, then a total of 4 different distribution methods: A: Twist, B: Twist, Bun A: Twist, twist, B: Bun A: Steamed buns, B: Twist, twist A: Twist, bun, B: Twist

Input data The first line is the number of students n and the number of specialties M. The second line contains m integers, representing the number of each specialty. N, M not exceeding 1000, the number of each specialty is not more than 1000

Output outputs a row with the total number of different allocation schemes. Since the output can be very large, you only need to output the value of MoD 1,000,000,007 for the final result.

Sample Input5 4
1 3 3 5Sample Output384835HINT Source

Mathematical problem Combination number tolerance

1#include <iostream>2#include <cstdio>3#include <algorithm>4#include <cstring>5 #defineLL Long Long6 using namespacestd;7 Const intmxn= .;8 Const intmod=1e9+7;9 intRead () {Ten     intx=0, f=1;CharCh=GetChar (); One      while(ch<'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();} A      while(ch>='0'&& ch<='9') {x=x*Ten-'0'+ch;ch=GetChar ();} -     returnx*F; - } the LL FAC[MXN],INV[MXN]; - voidinit () { -fac[0]=fac[1]=1; -inv[0]=inv[1]=1; +      for(intI=2; i<mxn;i++){ -fac[i]= (LL) fac[i-1]*i%MoD; +Inv[i]= ((-mod/i) *inv[mod%i]%mod+mod)%MoD; A     } at      for(intI=2; i<mxn;i++) -inv[i]= (LL) inv[i]*inv[i-1]%MoD; -     return; - } -LL Calc (intNintm) { -     if(!m)return 1; in     if(n<m)return 0; -     return(LL) fac[n]*inv[m]%mod*inv[n-m]%MoD; to } + intn,m; - inta[mxn],smm=0; the LL F[MXN]; * intMain () { $     inti,j;Panax Notoginseng init (); -N=read (); m=read (); the      for(i=1; i<=m;i++) +a[i]=read (); A      for(i=1; i<=n;i++){ thef[i]=1; +          for(j=1; j<=m;j++) -F[i]=f[i]*calc (I-1+A[J],A[J])%MoD; $          for(j=1; j<i;j++){ $F[i]= (F[i]-f[j]*calc (i,j)%mod+mod)%MoD; -         } -     } theprintf"%lld\n", F[n]); -     return 0;Wuyi } the 

Bzoj4710 [Jsoi2011] sub-specialty