Codeforces 425E Sereja and sets

Serie A champions:

Set S It consists of a number of intervals [l,r] and 1<=l<=r<=n F (s) that do not intersect the interval asking for N and F (s) There are several possible s sets

Ideas:

DP good question as to why is DP ... I can only say that the fat God taught me--b

Define DP[I][J] means that when N=i and F (s) =j the number of S collection species then it can be obtained by dp[k][j-1] J-1<=k<=i

To be able to understand this transfer

The first thing we need to do is to add a disjoint interval to j-1 J, [k+1,k+1] [k+1,k+2] ... Can all have a common 2^ (i-k)-1 kinds of extraction

1 because not all of this range is not taken

Why not [k+2,...] Because I'm the k+2 that traverses K. The interval that begins is counted when traversing to k=k+1.

And then we're able to take some intervals. These intervals must intersect with the previous ones.

How does it intersect? Just let one end be within K. Therefore, a co-owned 2^ ((i-k) *k) seed extraction

Final draw Dp[i][j] = Sigma (dp[k][j-1] * (2^ (I-K)-1) * (2^ ((i-k) * k))% MoD

Code:

#include <cstdio> #include <cstring> #include <algorithm> #include <vector>using namespace std; #define N 510#define LL __int64#define mod 1000000007LL dp[n][n],f[n*n];int n,m;int Main () {    int i,j,k;    scanf ("%d%d", &n,&m);    for (i=1,f[0]=1;i<n*n;i++) f[i]= (f[i-1]<<1)%mod;    for (i=0;i<=n;i++) dp[i][0]=1;    for (i=1;i<=n;i++)    {for (        j=1;j<=min (m,i); j + +)        {for            (k=j-1;k<=i;k++)            {                dp[i][j ]= (dp[i][j]+dp[k][j-1]* (f[i-k]-1)%mod*f[k* (i-k)]%mod)%mod;    }}} printf ("%i64d\n", Dp[n][m]);    return 0;}

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Codeforces 425E Sereja and sets