Luogu 4492 [HAOI2018] Number of apple tree combinations

Source: Internet
Author: User

To find the contribution of the parent side of each numbered point, the combination number and factorial can be calculated.

I didn't think of it in the examination room.

Adjusted for a long time and long a long, not to be able to tune out, the sample has been over, just found that overflow, I am a zz.

1#include <cstdio>2#include <cstring>3#include <algorithm>4#include <cmath>5#include <iostream>6 using namespacestd;7 #defineLL Long Long8 LL N; LL p;9LL f[ .][ .]={};TenLL zu[ .][ .]={}; OneLL t[ .]={}; A intMain () { -scanf"%lld%lld",&n,&p); -LL ans=0; zu[0][0]=1; t[0]=1; the      for(LL i=1; i<=n;i++) t[i]= (t[i-1]*i)%p; -      for(LL i=1; i<=n;i++){ -zu[i][0]=1; -          for(LL j=1; j<=i;j++) zu[i][j]= (zu[i-1][j]+zu[i-1][j-1])%p; +     } -      for(LL i=1; i<=n;i++){ +f[i][0]=1; A          for(intj=1; j<=n;j++){ atF[i][j]= (f[i][j-1]* (i+j-1))%p; -         } -     } -      for(LL i=2; i<=n;i++){ -          for(LL j=n-i+1;j>0; j--){ -LL w= (zu[n-i][j-1]*T[J])%p; inW= (w* (t[i]*f[i-1][n-i-j+1]) (%p))%p; -W= (w* ((j* (n-j))%p)%p; toans= (ans+w)%p; +         } -     } theprintf"%lld\n", ans); *     return 0; $}
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Luogu 4492 [HAOI2018] Number of apple tree combinations

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