[字符串Hash DP] Codeforces 613E #339 (Div. 1) E. Puzzle Lover

写这个东西的时候很需要冷静 一些细节 比如特殊情况不可避免的重复计算 很恶心
然后 自然溢出被sxbk的hacker卡了
我还因此去学习了怎么卡 构造串卡掉自然溢出BY PoPoQQQ

#include<cstdio>#include<cstdlib>#include<algorithm>#include<cstring>#define cl(x) memset(x,0,sizeof(x))using namespace std;typedef long long ll;inline char nc(){  static char buf[100000],*p1=buf,*p2=buf;  return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline int read(char *s){  char c=nc(); int len=0;  for (;!(c>='a' && c<='z');c=nc());  for (;c>='a' && c<='z';s[++len]=c,c=nc()); s[++len]=0; return len-1;}const int P=1e9+7;const int N=2005;inline void add(int &x,int y){  x+=y; if (x>=P) x-=P; }const ll S=31;const ll MOD=987654321;ll seed[N];inline void Pre(int n){  seed[0]=1;  for (int i=1;i<=n;i++)    seed[i]=seed[i-1]*S%MOD;}struct Hash{  ll h[N];  inline void make(int n,char *s){    h[0]=0;    for (int i=1;i<=n;i++)      h[i]=(h[i-1]*S+s[i]-'a')%MOD;  }  inline ll cut(int l,int r){    return (h[r]+MOD-h[l-1]*seed[r-l+1]%MOD)%MOD;  }}pre[2],suf[2],ss;int n,m;char a[2][N],s[N];int f[2][N][N];inline int Solve(int flag){  int ret=0; cl(f);  for (int j=1;j<=n;j++){    f[0][j][0]=f[1][j][0]=1;    for (int i=0;i<2;i++)      for (int k=2;k<=min(n-j+1,m/2);k++)    if (ss.cut(m-2*k+1,m-k)==pre[i].cut(j,j+k-1) && ss.cut(m-k+1,m)==suf[i^1].cut(n-(j+k-1)+1,n-j+1))      if (2*k!=m || flag)        add(ret,f[i][j][m-2*k]);    for (int i=0;i<2;i++)      for (int k=2;k<=min(j,m/2);k++)    if (ss.cut(k+1,2*k)==pre[i].cut(j-k+1,j) && ss.cut(1,k)==suf[i^1].cut(n-j+1,n-(j-k+1)+1))      if (2*k!=m || flag)        add(f[i][j+1][2*k],1);    for (int i=0;i<2;i++)      for (int k=0;k<m;k++)    if (a[i][j]==s[k+1]){      add(f[i][j+1][k+1],f[i][j][k]);      if (k+2<=m && a[i^1][j]==s[k+2])        add(f[i^1][j+1][k+2],f[i][j][k]);    }    for (int i=0;i<2;i++)      add(ret,f[i][j+1][m]);  }  return ret;}int main(){  freopen("string.in","r",stdin);  freopen("string.out","w",stdout);  Pre(2000);  n=read(a[0]); read(a[1]);  for (int i=0;i<2;i++){    pre[i].make(n,a[i]);    reverse(a[i]+1,a[i]+n+1);    suf[i].make(n,a[i]);    reverse(a[i]+1,a[i]+n+1);  }  m=read(s); ss.make(m,s);  int Ans=0;  add(Ans,Solve(1));  if (m>1){    reverse(s+1,s+m+1);    ss.make(m,s);    add(Ans,Solve(0));    if (m==2){      for (int j=1;j<=n;j++)    for (int i=0;i<2;i++)      if (a[i][j]==s[1] && a[i^1][j]==s[2])        add(Ans,P-1);    }  }  printf("%d\n",Ans);  return 0;}