bzoj 1223: [HNOI2002]Kathy函数 数位DP 高精度

1223: [HNOI2002]Kathy函数

Time Limit: 10 Sec  Memory Limit: 162 MB
Submit: 207  Solved: 90
[Submit][Status]

Description

Input

仅有一行，为正整数m

Output

输出仅有一个正整数，表示所有的满足f(n)=n,(n<=m) 的自然数的个数。

Sample Input

5

Sample Output

3

　　这道题的高精度模板相对又有完善，但还存在bugs，使用时尽量讲位数多开几倍。

　　这道题由于递归方程转移有2、4，所以应该往二进制那边去想，可以通过归纳法证明f(n)=n当且仅当n二进制为回文串（相当神奇），然后就是数位DP，最开始我将数据范围想成了m<=2^100，还说刚好long long可以存，但是10^100就必须写高精了。

　　

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<unistd.h>
using namespace std;
typedef long long qword;
#define MAXN 20100
class number//四位
{
        public:
                const static int max_val=10000;
                const static int max_len=100;
                number()
                {
                        clear();
                }
                number(int x)
                {
                        *this=x;
                }
                bool is_odd()const
                {
                        return numb[0]%2==1;
                }
                bool is_even()const
                {
                        return numb[0]%2==0;
                }
                int size()const
                {
                        return topn;
                }
                int length()const
                {
                        int x=numb[topn];
                        int ret=0;
                        while (x)
                        {
                                ret++;
                                x/=10;
                        }
                        int y=0;
                        x=number::max_val;
                        while (x)
                        {
                                y++;
                                x/=10;
                        }
                        y--;
                        ret+=topn*y;
                        return ret;
                }
                number pow(int x)const
                {
                        number ret,a;
                        a=*this;
                        ret=1;
                        while (x)
                        {
                                if (x&1)ret=ret*a;
                                a=a*a;
                                x>>=1;
                        }
                        return ret;
                }
                void operator =(int x)//{{{
                {
                        int now=0;
                        clear();
                        numb[now]=x;
                        while (numb[now]>=number::max_val)
                        {
                                numb[now+1]+=numb[now]/number::max_val;
                                numb[now]%=number::max_val;
                                now++;
                                if (now>topn)topn=now;
                        }
                }//}}}
void operator =(number num)
{
        topn=num.topn;
        memcpy((this->numb),num.numb,sizeof(num.numb[0])*(topn+1));
}
void operator +=(number num)//{{{
{
        int i;
        topn=max(topn,num.topn);
        for (i=0;i<=topn;i++)
        {
                numb[i]+=num.numb[i];;
                if (numb[i]>=number::max_val)
                {
                        numb[i+1]+=numb[i]/number::max_val;
                        numb[i]%=number::max_val;
                }
        }
        while (numb[topn+1])
        {
                topn++;
                numb[topn+1]+=numb[topn]/number::max_val;
                numb[topn]%=number::max_val;
        }
}//}}}
void operator +=(int x)//{{{
{
        int now=0;
        if (topn==-1)topn=0;
        numb[now]+=x;
        while (numb[now]>=number::max_val)
        {
                numb[now+1]+=numb[now]/number::max_val;
                numb[now]%=number::max_val;
                now++;
                if (now>topn)topn=now;
        }
}//}}}
void operator *=(int x)//{{{
{
        int i;
        for (i=0;i<=topn;i++)
        {
                numb[i]*=x;
        }
        for (i=0;i<=topn;i++)
        {
                if (numb[i]>=number::max_val)
                {
                        numb[i+1]+=numb[i]/number::max_val;
                        numb[i]%=number::max_val;
                }
        }
        while (numb[topn+1])
        {
                topn++;
                numb[topn+1]+=numb[topn]/number::max_val;
                numb[topn]%=number::max_val;
        }
}//}}}
void operator -=(number &num)//{{{
{
        if (*this<num)throw "Error!\n->void operator -=(number &num)\n";
        int i;
        for (i=0;i<=topn;i++)
        {
                numb[i]-=num.numb[i];
        }
        for (i=0;i<=topn;i++)
        {
                while (numb[i]<0)
                {
                        numb[i]+=number::max_val;
                        numb[i+1]--;
                }
        }
        while (topn&&!numb[topn])topn--;
}//}}}
void operator --(int)//{{{
{
        if (topn==0&&numb[0]==0)throw "Error!\n->void operator --(int)\n";
        int now=0;
        numb[now]--;
        while (numb[now]<0)
        {
                numb[now+1]--;
                numb[now]+=number::max_val;
        }
        while (topn&&!numb[topn])topn--;
}//}}}
private:
int numb[max_len];
int topn;
void clear()
{
        topn=0;
        memset(numb,0,sizeof(numb));

}
friend bool operator <(number num1,number num2);
friend bool operator <=(number num1,number num2);
friend bool operator ==(number num1,number num2);
friend ostream& operator <<(ostream &out,number num);
friend istream& operator >>(istream &in,number &num);
friend number operator *(number num1,number num2);
friend number operator *(number num,int x);
friend number operator +(number num1,number num2);
//a=a+b远没有a+=b快
};
bool operator <(number num1,number num2)//{{{
{
        if (num1.topn!=num2.topn)
        {
                return num1.topn<num2.topn;
        } 
        int i;
        for (i=num1.topn;i>=0;i--)
        {
                if (num1.numb[i]!=num2.numb[i])
                {
                        return num1.numb[i]<num2.numb[i];
                }
        }
        return false;
}//}}}
bool operator <=(number num1,number num2)//{{{
{
        if (num1.topn!=num2.topn)
        {
                return num1.topn<num2.topn;
        } 
        int i;
        for (i=num1.topn;i>=0;i--)
        {
                if (num1.numb[i]!=num2.numb[i])
                {
                        return num1.numb[i]<num2.numb[i];
                }
        }
        return true;
}//}}}
bool operator ==(number num1,number num2)//{{{
{
        if (num1.topn!=num2.topn)return false;
        for (int i=0;i<=num1.topn;i++)
        {
                if (num1.numb[i]!=num2.numb[i])return false;
        }
        return true;
}//}}}
ostream& operator <<(ostream &out,number num)//{{{
{
        int i;
        out<<num.numb[num.topn];
        for (i=num.topn-1;i>=0;i--)
        {
                //压六位时
                //if (num.numb[i]<100000)out<<"0";
                //if (num.numb[i]<10000)out<<"0";
                if (num.numb[i]<1000)out<<"0";
                if (num.numb[i]<100)out<<"0";
                if (num.numb[i]<10)out<<"0";
                out<<num.numb[i];
        }
        return out;
}//}}}
istream& operator >>(istream &in,number &num)//{{{
{
        string str;
        in>>str;
        int i;
        num.clear();
        for (i=(int)str.length()-1,num.topn=0;i>=0;i-=4,num.topn++)
        {
                if (i-3<(int)str.length())
                {
                        num.numb[num.topn]=(str[i]-'0')+10*(str[i-1]-'0')+100*(str[i-2]-'0')+1000*(str[i-3]-'0');
                }else
                {
                        if (i-2<(int)str.length())num.numb[num.topn]+=100*(str[i-2]-'0');
                        if (i-1<(int)str.length())num.numb[num.topn]+=10*(str[i-1]-'0');
                        if (i<(int)str.length())num.numb[num.topn]+=(str[i]-'0');
                }
        }
        num.topn--;
        return in;
}//}}}
number operator *(number num,int x)//{{{
{
        number ret;
        ret=num;
        ret*=x;
        return ret;
}//}}}
number operator *(number num1,number num2)
{
        number ret;
        ret.topn=num1.topn+num2.topn;
        for (int i=0;i<=num1.topn;i++)
                for (int j=0;j<=num2.topn;j++)
                {
                        ret.numb[i+j]+=num1.numb[i]*num2.numb[j];
                        ret.numb[i+j+1]+=ret.numb[i+j]/number::max_val;
                        ret.numb[i+j]%=number::max_val;
                }
        for (int i=0;i<=ret.topn;i++)
        {
                ret.numb[i+1]+=ret.numb[i]/number::max_val;
                ret.numb[i]%=number::max_val;
        }
        while (ret.numb[ret.topn+1])
        {
                ret.topn++;
                ret.numb[ret.topn+1]+=ret.numb[ret.topn]/number::max_val;
                ret.numb[ret.topn]%=number::max_val;
        }
        while (ret.topn && !ret.numb[ret.topn])
        {
                ret.topn--;
        }
        return ret;
}
number operator +(number num1,number num2)//{{{
{
        number ret;
        ret=num1;
        ret+=num2;
        return ret;
}//}}}

char str[MAXN];
int num0[MAXN];
bool num[MAXN];
number rec[MAXN][2][2];
int n,m;
number search_m(int now,int fl,int bo)
{
        int x,y;
        x=now-1;y=n-now;
        if (x>y)
        {
                number ret;
                ret=(int)(fl || (!fl && !bo));
                return ret;
        }
        number &res=rec[now][fl][bo];
        number aa;
        if (aa<rec[now][fl][bo])return res;
        res=0;
        if (!fl)
        {
                if (num[y])
                {
                        //1
                        if (num[x]==1)
                                res+=search_m(now+1,false,bo);//1
                        else
                                res+=search_m(now+1,false,true);//1

                        if (now>1)//首位非0
                        {
                                if (num[x]==1)
                                        res+=search_m(now+1,true,false);//0
                                else
                                        res+=search_m(now+1,true,bo);//0
                        }
                }else if (!num[y])
                {
                        if (now>1)//特殊处理m==0情况
                        {
                                if (num[x]==1)
                                        res+=search_m(now+1,false,false);//0
                                else
                                        res+=search_m(now+1,false,bo);//0
                        }
                }
        }else
        {
                res+=search_m(now+1,true,true);//1
                res+=search_m(now+1,true,true);//0
        }
        return res;
}
int main()
{
        //freopen("input.txt","r",stdin);
        scanf("%s\n",str);
        int i,j,x,y;
        n=strlen(str);
        x=y=-1;
        for (i=n-1;i>=0;i-=4)
        {
                x++;
                for (j=max(0,i-3);j<=i;j++)
                        num0[x]=num0[x]*10+str[j]-'0';
        }
        n=x+1;
        x=0;
        while (~n)
        {
                num[x++]=num0[0]%2;
                for (i=n-1;i>=0;i--)
                {
                        if (i)num0[i-1]+=(num0[i]&1)*10000;
                        num0[i]/=2;
                }
                while (n>=0 && !num0[n-1])n--;
        }
        n=x;
        /*for (i=n-1;i>=0;i--)
          {
          printf("%d",num[i]);
          }
          printf("\n");*/
        /*for (i=0;i<=n;i++)
          for (j=0;j<2;j++)
          for (k=0;k<2;k++)
          rec[i][j][k]=-1;*/
        number ans;
        ans=search_m(1,false,false);
        number xx;
        for (i=1;i<n;i++)
        {
                xx=2;
                xx=xx.pow((i-1)/2);
                ans+=xx;
        }
        cout<<ans<<endl;
        return 0;
}