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bzoj 4461: [Jsoi2013]美丽家园

2016-05-31 10:40 369 查看
千分矩阵乘法题

dp[i][j]表示第i行,j状态是否可行

矩乘就好,需要高精度

#include<bits/stdc++.h>
using namespace std;
inline void splay(int&v){
v=0;char c=0;int p=1;
while(c<'0'||c>'9'){if(c=='-')p=-1;c=getchar();}
while(c>='0'&&c<='9'){v=(v<<3)+(v<<1)+c-'0';c=getchar();}
v*=p;
}
const int MAXD = 12, DIG = 9, BASE = 1000000000;
const unsigned long long BOUND = numeric_limits <unsigned long long> :: max () - (unsigned long long) BASE * BASE;
class bignum{private:int digits[MAXD];int D;
public:friend ostream &operator<<(ostream &out,bignum &c);
inline void trim(){while(D>1&&digits[D-1]==0)D--;}
inline void dealint(long long x){memset(digits,0,sizeof(digits));D=0;do{digits[D++]=x%BASE;x/=BASE;}while(x>0);  }
inline void dealstr(char *s){memset(digits,0,sizeof(digits));int len=strlen(s),first=(len+DIG-1)%DIG+1;D=(len+DIG-1)/DIG;for(int i=0;i<first;i++)digits[D-1]=digits[D-1]*10+s[i]-'0';for(int i=first,d=D-2;i<len;i+=DIG,d--)for(int j=i;j<i+DIG;j++)digits[d]=digits[d]*10+s[j]-'0';trim();}
inline char *print(){
trim();char *cdigits=new char[DIG*D+1];int pos=0,d=digits[D-1];do{cdigits[pos++]=d%10+'0';d/=10;}while(d > 0);reverse(cdigits,cdigits+pos);for(int i=D-2;i>=0;i--,pos += DIG)
for(int j=DIG-1,t=digits[i];j>= 0;j--){cdigits[pos+j]=t%10+'0';t/=10;}cdigits[pos]='\0';return cdigits;}
bignum(){dealint(0);}
bignum(long long x){dealint(x);}
bignum(int x){dealint(x);}
bignum(char *s){dealstr(s);}
inline bool operator < (const bignum &o) const{if(D != o.D)return D < o.D;for(int i = D-1; i>=0; i--)if(digits[i] != o.digits[i])return digits[i] < o.digits[i];return false;}
bool operator >  (const bignum & o)const{return o < *this;}
bool operator <= (const bignum & o)const{return !(o < *this);}
bool operator >= (const bignum & o)const{return !(*this < o);}
bool operator != (const bignum & o)const{return o < *this || *this < o;}
bool operator == (const bignum & o)const{return !(o < *this) && !(*this < o);}
bignum &operator++(){*this = *this  + 1;return *this;}
bignum operator ++(int){bignum old = *this;++(*this);return old;}
inline bignum operator << (int p) const{bignum temp;temp.D=D+p;
for(int i=0;i<D;i++)temp.digits [i + p] = digits [i];for (int i = 0; i < p; i++)temp.digits [i] = 0;return temp;}
inline bignum operator >> (int p)const{bignum temp;temp.D=D-p;for(int i=0;i<D-p;i++)temp.digits[i]=digits[i+p];for(int i=D-p;i<D;i++)temp.digits[i]=0;return temp;}
bignum &operator += (const bignum &b){  *this = *this + b;  return *this;}
bignum &operator -= (const bignum &b){  *this = *this - b;  return *this;  }
bignum &operator *= (const bignum &b){  *this = *this * b;  return *this;  }
bignum &operator /= (const bignum &b){  *this = *this / b;  return *this;  }
bignum &operator %= (const bignum &b){  *this = *this % b;  return *this;  }
inline bignum operator + (const bignum &o) const  {  bignum sum = o;  int carry = 0;  for (sum.D = 0; sum.D < D || carry > 0; sum.D++)  {  sum.digits [sum.D] += (sum.D < D ? digits [sum.D] : 0) + carry;  if (sum.digits [sum.D] >= BASE)  {  sum.digits [sum.D] -= BASE;  carry = 1;  }  else  carry = 0;  }  sum.D = max (sum.D, o.D);  sum.trim ();  return sum;  }
inline bignum operator - (const bignum &o) const  {  bignum diff = *this;  for (int i = 0, carry = 0; i < o.D || carry > 0; i++)  {  diff.digits [i] -= (i < o.D ? o.digits [i] : 0) + carry;  if (diff.digits [i] < 0)  {  diff.digits [i] += BASE;  carry = 1;  }  else  carry = 0;  }  diff.trim ();  return diff;  }
inline bignum operator * (const bignum &o) const  {  bignum prod = 0;  unsigned long long sum = 0, carry = 0;  for (prod.D = 0; prod.D < D + o.D - 1 || carry > 0; prod.D++)  {  sum = carry % BASE;  carry /= BASE;  for (int j = max (prod.D - o.D + 1, 0); j <= min (D - 1, prod.D); j++)  {  sum += (unsigned long long) digits [j] * o.digits [prod.D - j];  if (sum >= BOUND)  {  carry += sum / BASE;  sum %= BASE;}}carry += sum / BASE;prod.digits [prod.D] = sum % BASE;}prod.trim ();return prod;}
inline bignum range (int a, int b) const{  bignum temp = 0;  temp.D = b - a;  for (int i = 0; i < temp.D; i++)  temp.digits [i] = digits [i + a];  return temp;  }
inline double double_div (const bignum &o) const  {  double val = 0, oval = 0;  int num = 0, onum = 0;  for (int i = D - 1; i >= max (D - 3, 0); i--, num++)  val = val * BASE + digits [i];  for (int i = o.D - 1; i >= max (o.D - 3, 0); i--, onum++)  oval = oval * BASE + o.digits [i];  return val / oval * (D - num > o.D - onum ? BASE : 1);  }
inline pair <bignum, bignum> divmod (const bignum &o) const  {  bignum quot = 0, rem = *this, temp;  for (int i = D - o.D; i >= 0; i--)  {  temp = rem.range (i, rem.D);  int div = (int) temp.double_div (o);  bignum mult = o * div;  while (div > 0 && temp < mult)  {  mult = mult - o;  div--;  }  while (div + 1 < BASE && !(temp < mult + o))  {  mult = mult + o;  div++;  }  rem = rem - (o * div << i);  if (div > 0)  {  quot.digits [i] = div;  quot.D = max (quot.D, i + 1);  }  }  quot.trim ();  rem.trim ();  return make_pair (quot, rem);  }
inline bignum operator / (const bignum &o) const  {  return divmod (o).first;  }
inline bignum operator % (const bignum &o) const  {  return divmod (o).second;  }
inline bignum power (int exp) const  {  bignum p = 1, temp = *this;  while (exp > 0)  {  if (exp & 1) p = p * temp;  if (exp > 1) temp = temp * temp;  exp >>= 1;  }  return p;  }
inline bignum factorial() const  {  bignum ans = 1, num = *this;  if (num == 0 || num == 1)  return ans;  while (!(num < 0 || num == 0))  {  ans = ans * num;  num = num - 1;  }  return ans;  }  };
ostream &operator<<(ostream &out, bignum &c)  {  out<<c.print();  return out;  }
istream &operator >> (istream &in,bignum &c)  {  char s[10000];  in>>s;  c = s;  return in;  }
bignum gcd(bignum a,bignum b){return b==0?a:gcd(b,a%b);}

bignum n;
int m,mod;
struct M{
unsigned v[33][33],f;
M(){
memset(v,0,sizeof v);f=0;
}
friend M operator * (const M &a,const M &b){
M c;
for(int i=1;i<=a.f;i++){
for(int j=1;j<=a.f;j++){
for(int k=1;k<=a.f;k++){
c.v[i][j]+=a.v[i][k]*b.v[k][j];
}
}
}
c.f=a.f;
for(int i=1;i<=c.f;i++){
for(int j=1;j<=c.f;j++){
c.v[i][j]%=mod;
}
}
return c;
}
}A,B,C;
int main(){
freopen("xxx.in","r",stdin);
freopen("xxx.out","w",stdout);
cin>>n>>m>>mod;
A.f=B.f=C.f=1<<m;
for(int i=1;i<=A.f;i++)A.v[i][i]=1;
for(int i=1;i<=B.f;i++){
for(int j=1;j<=B.f;j++){
int a=i-1,b=j-1,flag=1;
for(int k=1;k<m;k++){
if((a&b&3)==3)flag=0;
if(((a^63)&(b^63)&3)==3)flag=0;
a>>=1,b>>=1;
}
B.v[i][j]=flag;
}
}
n=n-1;
while(n!=0){
if(n%2==1)A=A*B;
n=n/2;B=B*B;
}
int ans=0;
for(int i=1;i<=A.f;i++){
for(int j=1;j<=A.f;j++){
ans+=A.v[i][j];
}
}
cout<<ans%mod<<endl;
}
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