bzoj 4461: [Jsoi2013]美丽家园
2024-04-30 15:21:38
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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