codeforces613E

Puzzle Lover

 CodeForces - 613E 

Oleg Petrov loves crossword puzzles and every Thursday he buys his favorite magazine with crosswords and other word puzzles. In the last magazine Oleg found a curious puzzle, and the magazine promised a valuable prize for it's solution. We give a formal description of the problem below.

The puzzle field consists of two rows, each row contains n cells. Each cell contains exactly one small English letter. You also are given a word w, which consists of ksmall English letters. A solution of the puzzle is a sequence of field cells c₁, ..., c_k, such that:

• For all i from 1 to k the letter written in the cell c_i matches the letter w_i;
• All the cells in the sequence are pairwise distinct;
• For all i from 1 to k - 1 cells c_i and c_i + 1 have a common side.

Oleg Petrov quickly found a solution for the puzzle. Now he wonders, how many distinct solutions are there for this puzzle. Oleg Petrov doesn't like too large numbers, so calculate the answer modulo 10⁹ + 7.

Two solutions c_i and c'_i are considered distinct if the sequences of cells do not match in at least one position, that is there is such j in range from 1 to k, such that c_j ≠ c'_j.

Input

The first two lines contain the state of the field for the puzzle. Each of these non-empty lines contains exactly n small English letters.

The next line is left empty.

The next line is non-empty and contains word w, consisting of small English letters.

The length of each line doesn't exceed 2 000.

Output

Print a single integer — the number of distinct solutions for the puzzle modulo 10⁹ + 7.

Examples

Input

code
edoc

code

Output

4

Input

aaa
aaa

aa

Output

14

sol：超好的一道dp题

 

事实上我代码和上面略有不同
提醒一下：注意一点就是不要算重，比如要向左或向右绕一圈回来的时候一圈的长度最小值为4，因为直接往下走的话在中间那段的转移中已经算过了

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
inline ll read()
{
    ll s=0;
    bool f=0;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<3)+(s<<1)+(ch^48); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0)
    {
        putchar('-'); x=-x;
    }
    if(x<10)
    {
        putchar(x+'0'); return;
    }
    write(x/10);
    putchar((x%10)+'0');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')

const int N=2005;
const ll Base=97,Mod=1000000007;
int n,Len;
char S[2][N],T[N];
inline void Ad(ll &x,ll y)
{
    x+=y; x-=(x>=Mod)?Mod:0;
}
ll Seed[N];
inline void Prepare(int n)
{
    int i; Seed[0]=1;
    for(i=1;i<=n;i++) Seed[i]=1ll*Seed[i-1]*Base%Mod;
}
struct Hash
{
    ll Hash[N];
    inline void Make(int n,char *S)
    {
        int i; Hash[0]=0; for(i=1;i<=n;i++) Hash[i]=(Hash[i-1]*Base%Mod+(S[i]-'a'+1))%Mod;
    }
    inline ll Ask(int l,int r)
    {
        return (ll)(Hash[r]+Mod-Hash[l-1]*Seed[r-l+1]%Mod)%Mod;
    }
}Pre[2],Suf[2],HT;
ll dp[2][N][N];
//dp[i][j][k]表示在第i行,第j-1个(即i,j在当前点右边),已经匹配了k为的方案数 
inline ll Solve(bool opt)
{
    ll res=0;
    int i,j,k;
    memset(dp,0,sizeof dp);
    for(j=1;j<=n;j++)
    {
        dp[0][j][0]=dp[1][j][0]=1;
        for(i=0;i<=1;i++) for(k=2;k<=min(n-j+1,Len/2);k++) //向右绕一圈回来 
        {
            if(Pre[i].Ask(j,j+k-1)==HT.Ask(Len-2*k+1,Len-k)&&Suf[i^1].Ask(n-(j+k-1)+1,n-j+1)==HT.Ask(Len-k+1,Len))
            {
                if((k*2!=Len)||opt) Ad(res,dp[i][j][Len-k*2]);
            }
        }
        for(i=0;i<=1;i++) for(k=2;k<=min(j,Len/2);k++) //向左绕一圈回来 
        {
            if(Suf[i].Ask(n-j+1,n-(j-k+1)+1)==HT.Ask(1,k)&&Pre[i^1].Ask(j-k+1,j)==HT.Ask(k+1,k*2))
            {
                if((k*2!=Len)||opt) Ad(dp[i^1][j+1][k*2],1);
            }
        }
        for(i=0;i<=1;i++) for(k=1;k<=Len;k++)
        {
            if(T[k]==S[i][j])
            {
                Ad(dp[i][j+1][k],dp[i][j][k-1]);
                if(k<Len&&T[k+1]==S[i^1][j])
                {
                    Ad(dp[i^1][j+1][k+1],dp[i][j][k-1]);            
                }
            }
        }
        for(i=0;i<=1;i++) Ad(res,dp[i][j+1][Len]);
    }
    return res;
}
int main()
{
    ll i,j,ans=0;
    Prepare(2000);
    scanf("%s",S[0]+1); n=strlen(S[0]+1); scanf("%s",S[1]+1); scanf("%s",T+1); Len=strlen(T+1);
    for(i=0;i<=1;i++)
    {
        Pre[i].Make(n,S[i]); reverse(S[i]+1,S[i]+n+1); Suf[i].Make(n,S[i]); reverse(S[i]+1,S[i]+n+1);
    }
    HT.Make(Len,T);
    Ad(ans,Solve(1));
    if(Len>1)
    {
        reverse(T+1,T+Len+1); HT.Make(Len,T); Ad(ans,Solve(0));
    }
    if(Len==2)
    {
        for(j=1;j<=n;j++) for(i=0;i<=1;i++) if(S[i][j]==T[1]&&S[i^1][j]==T[2]) Ad(ans,Mod-1);
    }
    Wl(ans);
    return 0;
}
/*
input
code
edoc
code
output
4

input
aaa
aaa
aa
output
14

input
v
s
sv
output
1
*/