51nod-1296: a restricted arrangement

"Portal: 51nod-1296" Brief test Instructions:

There is a collection of 1 to N of the number in the collection

Given a constraint, A[i] indicates that the number of positions in the a[i] is smaller than the number of adjacent positions

Given the b constraint, B[i] indicates that the number of positions in the b[i] is greater than the number of adjacent positions

Find the number of sequences that match a condition

The following:

Dp

Set F[I][J] is the i digit, the last one is the number of cases of J

We have to put J reasonable at the end, so we can put the number of the front >=j +1, so we can empty the number of J.

Set P[i] for the first position of the constraints, P[i]=-1 indicates that the first position is less than the i-1 position, P[i]=1 is greater than, p[i]=0 means no restriction conditions

In order to be able to better transfer, set $s[j]=\sum_{k=1}^{j}f[i-1][k]$

Can be transferred in three kinds of situations

Reference Code:

#include <cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<algorithm>using namespaceStd;typedefLong LongLL;intp[5100];intf[5100][5100],s[5100];intmod=1e9+7;intMain () {intn,k,l; scanf ("%d%d%d",&n,&k,&l); BOOLbk=false;  for(intI=1; i<=k;i++)    {        intx; scanf ("%d", &x); x + +; if(p[x]==1) bk=true; P[X]=-1; if(x+1>n)Continue; if(p[x+1]==-1) bk=true; P[x+1]=1; }     for(intI=1; i<=l;i++)    {        intx; scanf ("%d", &x); x + +; if(p[x]==-1) bk=true; P[X]=1; if(x+1>n)Continue; if(p[x+1]==1) bk=true; P[x+1]=-1; }    if(bk==true) {printf ("0\n");return 0;} f[1][1]=1;  for(intI=2; i<=n;i++)    {         for(intj=1; j<i;j++) s[j]= (LL) s[j-1]+ (LL) f[i-1][J])%Mod;  for(intj=1; j<=i;j++)        {            if(p[i]==1) f[i][j]=s[j-1]; Else if(p[i]==-1) f[i][j]= (LL) s[i-1]-(LL) s[j-1]+ (LL) Mod)%Mod; Elsef[i][j]=s[i-1]; }    }    intans=0;  for(intI=1; i<=n;i++) ans= (LL) ans+ (LL) f[n][i])%Mod; printf ("%d\n", ans); return 0;}

51nod-1296: a restricted arrangement