Description

Input

Output

Sample Input

Sample Output

HINT

樣例即為文中(圖3)。共有兩種不同的輸出序列形式，序列BAB有1種產生方式，而序列BBA有2種產生方式，因此答案為5。 【大致數據規模】約30%的數據滿足 n, m ≤ 12； 約100%的數據滿足n, m ≤ 500。

Source

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#define N 503#define p 1024523using namespace std;int n,m,f[N][N][N];char s[N],s1[N];int main(){	freopen("a.in","r",stdin);    scanf("%d%d",&n,&m);    scanf("%s",s+1);    scanf("%s",s1+1);    //f[0][0][0]=1;    for (int i=0;i<=n;i++)     for (int j=0;j<=m;j++)      for (int k=0;k<=n;k++) {      	if (i==0&&j==0&&k==0) {      		f[0][0][0]=1;      		break;		  }      	int l=i+j-k;      	if (l<0) break;      	if (i-1>=0&&k-1>=0&&s[i]==s[k]) f[i][j][k]=(f[i][j][k]+f[i-1][j][k-1])%p;      	if (i-1>=0&&l-1>=0&&s[i]==s1[l]) f[i][j][k]=(f[i][j][k]+f[i-1][j][k])%p;      	if (j-1>=0&&k-1>=0&&s1[j]==s[k]) f[i][j][k]=(f[i][j][k]+f[i][j-1][k-1])%p;      	if (j-1>=0&&l-1>=0&&s1[j]==s1[l]) f[i][j][k]=(f[i][j][k]+f[i][j-1][k])%p;	  }    PRintf("%d/n",f[n][m][n]);}