[期望 DP || 高斯消元 KMP] BZOJ 3213 [Zjoi2013]抛硬币

这个其实也不复杂 

先kmp 可以发现 每个点的状态会从转移到 i+1 和 next[i] 不妨设为f

然后列出方程 直接就可以上高斯消元 大概80？

这个东西其实可以DP

E(i+1)=(E(i)-1-p[t^1]*E(f))/p[t]

E(i)=k[i]*E(0)-b[i]

k[i+1]*E(i+1)-b[i+1]=(k[i]*E(0)-b[i]-1-p[t^1]*(k[f]*E(0)-b[f]))/p[t]
k[i+1]=(k[i]-p[t^1]*k[F[i][t^1]])/p[t]
b[i+1]=(b[i]+1-p[t^1]*b[F[i][t^1]])/p[t]

然后最后用E(n)==0就解出来了

只是为什么k b这么设呢 

可以发现 k b 这样的话已知非负 

甚至可以发现 k 一直为1 因为 k0=1 k1=1 ... 然后一直归纳下去就好了

这样就只要dp b数组就行了 

#include<cstdio>#include<cstdlib>#include<algorithm>#include<cstring>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 int read(char *s){  char c=nc(); int len=0;  for (;!(c=='H' || c=='T');c=nc());  for (;c=='H' || c=='T';s[++len]=c,c=nc()); s[++len]=0; return len-1;}const int con=100000000;class Int{public:  long long a[305];  Int() { }  Int(int x){memset(a,0,sizeof(a));while (x){a[++a[0]]=x%con;x=x/con;}}  void print(){    if (a[0]==0||(a[0]==1&&a[1]==0)){printf("0");return;}    printf("%lld",a[a[0]]); for (int i=a[0]-1;i;i--) printf("%08lld",a[i]);  }  Int operator +(const Int &X){    Int c;memset(c.a,0,sizeof(c.a));    for (int i=1;i<=a[0]||i<=X.a[0];i++) c.a[i]=c.a[i]+a[i]+X.a[i],c.a[i+1]+=c.a[i]/con,c.a[i]%=con;    c.a[0]=max(a[0],X.a[0]); if (c.a[c.a[0]+1])c.a[0]++; return c;  }  Int operator -(const Int &X){    Int c;memcpy(c.a,a,sizeof(c.a));    for (int i=1;i<=a[0];i++){ c.a[i]=c.a[i]-X.a[i]; if (c.a[i]<0) c.a[i+1]--,c.a[i]+=con; }    while (c.a[0]&&!c.a[c.a[0]]) c.a[0]--; return c;  }  Int operator *(const Int &X){    Int c;memset(c.a,0,sizeof(c.a));    for (int i=1;i<=a[0];i++)      for (int j=1;j<=X.a[0];j++)c.a[i+j-1]+=a[i]*X.a[j],c.a[i+j]+=c.a[i+j-1]/con,c.a[i+j-1]%=con;    c.a[0]=max(a[0]+X.a[0]-1,0ll); if (c.a[a[0]+X.a[0]]>0)c.a[0]++; return c;  }  Int operator *(int num){    Int c;memset(c.a,0,sizeof(c.a));    for (int i=1;i<=a[0];i++){ c.a[i]+=a[i]*num; if (c.a[i]>=con) c.a[i+1]+=c.a[i]/con,c.a[i]%=con; }    c.a[0]=a[0]; if (c.a[c.a[0]+1]>0)c.a[0]++; return c;  }  Int operator /(int num){    Int c;memset(c.a,0,sizeof(c.a));    long long x=0; for (int i=a[0];i;i--) x=x*con+a[i],c.a[i]=x/num,x=x%num;    c.a[0]=a[0]; if (c.a[0]&&!c.a[c.a[0]])c.a[0]--; return c;  }  long long operator %(int num){    Int c;memset(c.a,0,sizeof(c.a));    long long x=0; for (int i=a[0];i;i--) x=x*con+a[i],c.a[i]=x/num,x=x%num;    c.a[0]=a[0]; if (c.a[0]&&!c.a[c.a[0]])c.a[0]--; return x;  }};const int N=1005;struct frac{  Int a,b; // a/b  void doit(){ for (int i=2;i<=100;i++) while (b%i==0 && a%i==0) a=a/i,b=b/i; }  frac() {}  frac(int a,int b):a(a),b(b){ doit(); }  frac(Int ia,Int ib){ a=ia; b=ib; }  void print(){ a.print(),putchar('/'),b.print(),putchar('\n'); }  friend frac operator + (frac A,int p){ return frac(A.a+A.b*p,A.b); }  friend frac operator + (frac A,frac B){ return frac(A.a*B.b+A.b*B.a,A.b*B.b); }  friend frac operator - (frac A,frac B){ return frac(A.a*B.b-A.b*B.a,A.b*B.b); }  friend frac operator * (frac A,frac B){ return frac(A.a*B.a,A.b*B.b); }  friend frac operator / (frac A,frac B){ return frac(A.a*B.b,A.b*B.a); }}p[2],k[N],b[N],Ans;int ia,ib,n;char S[N];int next[N],fail[N][2];inline void KMP(){  next[1]=0; int k=0;  for (int i=2;i<=n;i++){    while (k && S[i]!=S[k+1]) k=next[k];    if (S[i]==S[k+1]) k++;    next[i]=k;  }}int main(){  freopen("coin.in","r",stdin);  freopen("coin.out","w",stdout);  read(ia); read(ib);  n=read(S); KMP();  p[0]=frac(ia,ib); p[1]=frac(ib-ia,ib);  int t=S[1]=='T',f; fail[1][t]=2; fail[1][t^1]=1;  for (int i=2;i<=n;i++)    t=S[i]=='T',fail[i][t]=i+1,fail[i][t^1]=fail[next[i-1]+1][t^1];  //k[1]=frac(1,1);  b[1]=frac(0,1);  for (int i=1;i<=n;i++){    t=S[i]=='T',f=fail[i][t^1];    //(k[i+1]=(k[i]-p[t^1]*k[f])/p[t]).doit();    (b[i+1]=(b[i]+frac(1,1)-p[t^1]*b[f])/p[t]).doit();    //k[i+1].print(); b[i+1].print();  }  //(Ans=b[n+1]/k[n+1]).doit();  (Ans=b[n+1]).doit();  Ans.print();  return 0;}