[平方的拆分 DP] BZOJ 1566 [NOI2009]管道取珠

根据乘法原理 方案数的平方等价于 操作两次结果相同的产生一的贡献

#include<cstdio>
#include<cstdlib>
#include<algorithm>
using namespace std;

inline char nc(){
static char buf[100000],*p1=buf,*p2=buf;
if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
return *p1++;
}

inline void read(int &x){
char c=nc(),b=1;
for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}

inline void read(char *s){
char c=nc(); int len=0;
for (;!(c>='A' && c<='B');c=nc());
for (;c>='A' && c<='B';s[++len]=c,c=nc()); s[++len]=0;
}

const int P=1024523;
const int N=505;

int n,m;
char A
,B
;
int f

;

inline void add(int &a,int b){
a+=b; if (a>P) a-=P;
}

int main()
{
read(n); read(m);
read(A); read(B);
reverse(A+1,A+n+1); reverse(B+1,B+m+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++)
{
int t=i+j-k;
if (i+1<=n && k+1<=n && A[i+1]==A[k+1])
add(f[i+1][j][k+1],f[i][j][k]);
if (i+1<=n && t+1<=m && A[i+1]==B[t+1])
add(f[i+1][j][k],f[i][j][k]);
if (j+1<=m && k+1<=n && B[j+1]==A[k+1])
add(f[i][j+1][k+1],f[i][j][k]);
if (j+1<=m && t+1<=m && B[j+1]==B[t+1])
add(f[i][j+1][k],f[i][j][k]);
}
printf("%d\n",f
[m]
);
return 0;
}