Acdream 1210 Chinese Girls' Amusement（大数模板运算 + 找规律）

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Chinese Girls’ Amusement
Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)
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Problem Description

  You must have heard that the Chinese culture is quite different from that of Europe or Russia. So some Chinese habits seem quite unusual or even weird to us.
  So it is known that there is one popular game of Chinese girls. N girls stand forming a circle and throw a ball to each other. First girl holding a ball throws it to the K-th girl on her left (1 ≤ K ≤ N/2). That girl catches the ball and in turn throws it to the K-th girl on her left, and so on. So the ball is passed from one girl to another until it comes back to the first girl. If for example N = 7 and K = 3, the girls receive the ball in the following order: 1, 4, 7, 3, 6, 2, 5, 1.

To make the game even more interesting the girls want to choose K as large as possible, but they want one condition to hold: each girl must own the ball during the game.
Input

Input contains one integer number N (3 ≤ N ≤ 102000) — the number of Chinese girls taking part in the game.
Output

Output the only number — K that they should choose.
Sample Input

7
6
Sample Output

3
1

题目大意：
有N(1　<　N　<　10^2000)个女孩围成一个圈从第1号女孩开始有一个球，可以往编号大的抛去（像传绣球一样绕着环来传），每次必须抛给左边第k个人，比如1号会抛给1+k号女孩。给出女孩的人数，如果他们都每个人都想要碰到球一次，那么这个k应该是多少（满足 1 ≤ K ≤ N/2 且 k必须尽量大）

解题思路：
其实我们就是从N/2开始倒着找GCD(x,N) == 1的那个数，其实一看这么大肯定是有规律的，所以可以先进行打个表找一下规律，其实我们发现当N是奇数的时候答案就是Ｎ/2，当N是偶数的时候，在进行判断如果Ｎ／２是偶数的话就是Ｎ／２－１，如果是奇数的话就是Ｎ／２－２，就是这个规律，然后套一个大数模板就OK啦（PS：我的这个大数模板很长但是可以当正常的ｉｎｔ用，建议可以参考，当正式比赛的时候，还是用一个比较省事儿的模板）
上代码：

#include <iostream>
#include <cstring>
using namespace std;

#define DIGIT   4      //四位隔开,即万进制
#define DEPTH   10000        //万进制
#define MAX     2000+5    //题目最大位数/4，要不大直接设为最大位数也行
typedef int bignum_t[MAX+1];

/************************************************************************/
/* 读取操作数，对操作数进行处理存储在数组里                             */
/************************************************************************/
int read(bignum_t a,istream&is=cin)
{
    char buf[MAX*DIGIT+1],ch ;
    int i,j ;
    memset((void*)a,0,sizeof(bignum_t));
    if(!(is>>buf))return 0 ;
    for(a[0]=strlen(buf),i=a[0]/2-1;i>=0;i--)
    ch=buf[i],buf[i]=buf[a[0]-1-i],buf[a[0]-1-i]=ch ;
    for(a[0]=(a[0]+DIGIT-1)/DIGIT,j=strlen(buf);j<a[0]*DIGIT;buf[j++]=‘0‘);
    for(i=1;i<=a[0];i++)
    for(a[i]=0,j=0;j<DIGIT;j++)
    a[i]=a[i]*10+buf[i*DIGIT-1-j]-‘0‘ ;
    for(;!a[a[0]]&&a[0]>1;a[0]--);
    return 1 ;
}

void write(const bignum_t a,ostream&os=cout)
{
    int i,j ;
    for(os<<a[i=a[0]],i--;i;i--)
    for(j=DEPTH/10;j;j/=10)
    os<<a[i]/j%10 ;
}

int comp(const bignum_t a,const bignum_t b)
{
    int i ;
    if(a[0]!=b[0])
    return a[0]-b[0];
    for(i=a[0];i;i--)
    if(a[i]!=b[i])
    return a[i]-b[i];
    return 0 ;
}

int comp(const bignum_t a,const int b)
{
    int c[12]=
    {
        1
    }
    ;
    for(c[1]=b;c[c[0]]>=DEPTH;c[c[0]+1]=c[c[0]]/DEPTH,c[c[0]]%=DEPTH,c[0]++);
    return comp(a,c);
}

int comp(const bignum_t a,const int c,const int d,const bignum_t b)
{
    int i,t=0,O=-DEPTH*2 ;
    if(b[0]-a[0]<d&&c)
    return 1 ;
    for(i=b[0];i>d;i--)
    {
        t=t*DEPTH+a[i-d]*c-b[i];
        if(t>0)return 1 ;
        if(t<O)return 0 ;
    }
    for(i=d;i;i--)
    {
        t=t*DEPTH-b[i];
        if(t>0)return 1 ;
        if(t<O)return 0 ;
    }
    return t>0 ;
}
/************************************************************************/
/* 大数与大数相加                                                       */
/************************************************************************/
void add(bignum_t a,const bignum_t b)
{
    int i ;
    for(i=1;i<=b[0];i++)
    if((a[i]+=b[i])>=DEPTH)
    a[i]-=DEPTH,a[i+1]++;
    if(b[0]>=a[0])
    a[0]=b[0];
    else
    for(;a[i]>=DEPTH&&i<a[0];a[i]-=DEPTH,i++,a[i]++);
    a[0]+=(a[a[0]+1]>0);
}
/************************************************************************/
/* 大数与小数相加                                                       */
/************************************************************************/
void add(bignum_t a,const int b)
{
    int i=1 ;
    for(a[1]+=b;a[i]>=DEPTH&&i<a[0];a[i+1]+=a[i]/DEPTH,a[i]%=DEPTH,i++);
    for(;a[a[0]]>=DEPTH;a[a[0]+1]=a[a[0]]/DEPTH,a[a[0]]%=DEPTH,a[0]++);
}
/************************************************************************/
/* 大数相减(被减数>=减数)                                               */
/************************************************************************/
void sub(bignum_t a,const bignum_t b)
{
    int i ;
    for(i=1;i<=b[0];i++)
    if((a[i]-=b[i])<0)
    a[i+1]--,a[i]+=DEPTH ;
    for(;a[i]<0;a[i]+=DEPTH,i++,a[i]--);
    for(;!a[a[0]]&&a[0]>1;a[0]--);
}
/************************************************************************/
/* 大数减去小数(被减数>=减数)                                           */
/************************************************************************/
void sub(bignum_t a,const int b)
{
    int i=1 ;
    for(a[1]-=b;a[i]<0;a[i+1]+=(a[i]-DEPTH+1)/DEPTH,a[i]-=(a[i]-DEPTH+1)/DEPTH*DEPTH,i++);
    for(;!a[a[0]]&&a[0]>1;a[0]--);
}

void sub(bignum_t a,const bignum_t b,const int c,const int d)
{
    int i,O=b[0]+d ;
    for(i=1+d;i<=O;i++)
    if((a[i]-=b[i-d]*c)<0)
    a[i+1]+=(a[i]-DEPTH+1)/DEPTH,a[i]-=(a[i]-DEPTH+1)/DEPTH*DEPTH ;
    for(;a[i]<0;a[i+1]+=(a[i]-DEPTH+1)/DEPTH,a[i]-=(a[i]-DEPTH+1)/DEPTH*DEPTH,i++);
    for(;!a[a[0]]&&a[0]>1;a[0]--);
}
/************************************************************************/
/* 大数相乘，读入被乘数a，乘数b，结果保存在c[]                          */
/************************************************************************/
void mul(bignum_t c,const bignum_t a,const bignum_t b)
{
    int i,j ;
    memset((void*)c,0,sizeof(bignum_t));
    for(c[0]=a[0]+b[0]-1,i=1;i<=a[0];i++)
    for(j=1;j<=b[0];j++)
    if((c[i+j-1]+=a[i]*b[j])>=DEPTH)
    c[i+j]+=c[i+j-1]/DEPTH,c[i+j-1]%=DEPTH ;
    for(c[0]+=(c[c[0]+1]>0);!c[c[0]]&&c[0]>1;c[0]--);
}
/************************************************************************/
/* 大数乘以小数，读入被乘数a，乘数b，结果保存在被乘数                   */
/************************************************************************/
void mul(bignum_t a,const int b)
{
    int i ;
    for(a[1]*=b,i=2;i<=a[0];i++)
    {
        a[i]*=b ;
        if(a[i-1]>=DEPTH)
        a[i]+=a[i-1]/DEPTH,a[i-1]%=DEPTH ;
    }
    for(;a[a[0]]>=DEPTH;a[a[0]+1]=a[a[0]]/DEPTH,a[a[0]]%=DEPTH,a[0]++);
    for(;!a[a[0]]&&a[0]>1;a[0]--);
}

void mul(bignum_t b,const bignum_t a,const int c,const int d)
{
    int i ;
    memset((void*)b,0,sizeof(bignum_t));
    for(b[0]=a[0]+d,i=d+1;i<=b[0];i++)
    if((b[i]+=a[i-d]*c)>=DEPTH)
    b[i+1]+=b[i]/DEPTH,b[i]%=DEPTH ;
    for(;b[b[0]+1];b[0]++,b[b[0]+1]=b[b[0]]/DEPTH,b[b[0]]%=DEPTH);
    for(;!b[b[0]]&&b[0]>1;b[0]--);
}
/**************************************************************************/
/* 大数相除,读入被除数a，除数b，结果保存在c[]数组                         */
/* 需要comp()函数                                                         */
/**************************************************************************/
void div(bignum_t c,bignum_t a,const bignum_t b)
{
    int h,l,m,i ;
    memset((void*)c,0,sizeof(bignum_t));
    c[0]=(b[0]<a[0]+1)?(a[0]-b[0]+2):1 ;
    for(i=c[0];i;sub(a,b,c[i]=m,i-1),i--)
    for(h=DEPTH-1,l=0,m=(h+l+1)>>1;h>l;m=(h+l+1)>>1)
    if(comp(b,m,i-1,a))h=m-1 ;
    else l=m ;
    for(;!c[c[0]]&&c[0]>1;c[0]--);
    c[0]=c[0]>1?c[0]:1 ;
}

void div(bignum_t a,const int b,int&c)
{
    int i ;
    for(c=0,i=a[0];i;c=c*DEPTH+a[i],a[i]=c/b,c%=b,i--);
    for(;!a[a[0]]&&a[0]>1;a[0]--);
}
/************************************************************************/
/* 大数平方根，读入大数a，结果保存在b[]数组里                           */
/* 需要comp()函数                                                       */
/************************************************************************/
void sqrt(bignum_t b,bignum_t a)
{
    int h,l,m,i ;
    memset((void*)b,0,sizeof(bignum_t));
    for(i=b[0]=(a[0]+1)>>1;i;sub(a,b,m,i-1),b[i]+=m,i--)
    for(h=DEPTH-1,l=0,b[i]=m=(h+l+1)>>1;h>l;b[i]=m=(h+l+1)>>1)
    if(comp(b,m,i-1,a))h=m-1 ;
    else l=m ;
    for(;!b[b[0]]&&b[0]>1;b[0]--);
    for(i=1;i<=b[0];b[i++]>>=1);
}
/************************************************************************/
/* 返回大数的长度                                                       */
/************************************************************************/
int length(const bignum_t a)
{
    int t,ret ;
    for(ret=(a[0]-1)*DIGIT,t=a[a[0]];t;t/=10,ret++);
    return ret>0?ret:1 ;
}
/************************************************************************/
/* 返回指定位置的数字，从低位开始数到第b位，返回b位上的数               */
/************************************************************************/
int digit(const bignum_t a,const int b)
{
    int i,ret ;
    for(ret=a[(b-1)/DIGIT+1],i=(b-1)%DIGIT;i;ret/=10,i--);
    return ret%10 ;
}
/************************************************************************/
/* 返回大数末尾0的个数                                                  */
/************************************************************************/
int zeronum(const bignum_t a)
{
    int ret,t ;
    for(ret=0;!a[ret+1];ret++);
    for(t=a[ret+1],ret*=DIGIT;!(t%10);t/=10,ret++);
    return ret ;
}

void comp(int*a,const int l,const int h,const int d)
{
    int i,j,t ;
    for(i=l;i<=h;i++)
    for(t=i,j=2;t>1;j++)
    while(!(t%j))
    a[j]+=d,t/=j ;
}

void convert(int*a,const int h,bignum_t b)
{
    int i,j,t=1 ;
    memset(b,0,sizeof(bignum_t));
    for(b[0]=b[1]=1,i=2;i<=h;i++)
    if(a[i])
    for(j=a[i];j;t*=i,j--)
    if(t*i>DEPTH)
    mul(b,t),t=1 ;
    mul(b,t);
}
/************************************************************************/
/* 组合数                                                               */
/************************************************************************/
void combination(bignum_t a,int m,int n)
{
    int*t=new int[m+1];
    memset((void*)t,0,sizeof(int)*(m+1));
    comp(t,n+1,m,1);
    comp(t,2,m-n,-1);
    convert(t,m,a);
    delete[]t ;
}
/************************************************************************/
/* 排列数                                                               */
/************************************************************************/
void permutation(bignum_t a,int m,int n)
{
    int i,t=1 ;
    memset(a,0,sizeof(bignum_t));
    a[0]=a[1]=1 ;
    for(i=m-n+1;i<=m;t*=i++)
    if(t*i>DEPTH)
    mul(a,t),t=1 ;
    mul(a,t);
}

#define SGN(x) ((x)>0?1:((x)<0?-1:0))
#define ABS(x) ((x)>0?(x):-(x))

int read(bignum_t a,int&sgn,istream&is=cin)
{
    char str[MAX*DIGIT+2],ch,*buf ;
    int i,j ;
    memset((void*)a,0,sizeof(bignum_t));
    if(!(is>>str))return 0 ;
    buf=str,sgn=1 ;
    if(*buf==‘-‘)sgn=-1,buf++;
    for(a[0]=strlen(buf),i=a[0]/2-1;i>=0;i--)
    ch=buf[i],buf[i]=buf[a[0]-1-i],buf[a[0]-1-i]=ch ;
    for(a[0]=(a[0]+DIGIT-1)/DIGIT,j=strlen(buf);j<a[0]*DIGIT;buf[j++]=‘0‘);
    for(i=1;i<=a[0];i++)
    for(a[i]=0,j=0;j<DIGIT;j++)
    a[i]=a[i]*10+buf[i*DIGIT-1-j]-‘0‘ ;
    for(;!a[a[0]]&&a[0]>1;a[0]--);
    if(a[0]==1&&!a[1])sgn=0 ;
    return 1 ;
}
struct bignum
{
    bignum_t num ;
    int sgn ;
    public :
    inline bignum()
    {
        memset(num,0,sizeof(bignum_t));
        num[0]=1 ;
        sgn=0 ;
    }
    inline int operator!()
    {
        return num[0]==1&&!num[1];
    }
    inline bignum&operator=(const bignum&a)
    {
        memcpy(num,a.num,sizeof(bignum_t));
        sgn=a.sgn ;
        return*this ;
    }
    inline bignum&operator=(const int a)
    {
        memset(num,0,sizeof(bignum_t));
        num[0]=1 ;
        sgn=SGN (a);
        add(num,sgn*a);
        return*this ;
    }
    ;
    inline bignum&operator+=(const bignum&a)
    {
        if(sgn==a.sgn)add(num,a.num);
        else if
        (sgn&&a.sgn)
        {
            int ret=comp(num,a.num);
            if(ret>0)sub(num,a.num);
            else if(ret<0)
            {
                bignum_t t ;
                memcpy(t,num,sizeof(bignum_t));
                memcpy(num,a.num,sizeof(bignum_t));
                sub (num,t);
                sgn=a.sgn ;
            }
            else memset(num,0,sizeof(bignum_t)),num[0]=1,sgn=0 ;
        }
        else if(!sgn)
            memcpy(num,a.num,sizeof(bignum_t)),sgn=a.sgn ;
        return*this ;
    }
    inline bignum&operator+=(const int a)
    {
        if(sgn*a>0)add(num,ABS(a));
        else if(sgn&&a)
        {
            int  ret=comp(num,ABS(a));
            if(ret>0)sub(num,ABS(a));
            else if(ret<0)
            {
                bignum_t t ;
                memcpy(t,num,sizeof(bignum_t));
                memset(num,0,sizeof(bignum_t));
                num[0]=1 ;
                add(num,ABS (a));
                sgn=-sgn ;
                sub(num,t);
            }
            else memset(num,0,sizeof(bignum_t)),num[0]=1,sgn=0 ;
        }
        else if
            (!sgn)sgn=SGN(a),add(num,ABS(a));
        return*this ;
    }
    inline bignum operator+(const bignum&a)
    {
        bignum ret ;
        memcpy(ret.num,num,sizeof (bignum_t));
        ret.sgn=sgn ;
        ret+=a ;
        return ret ;
    }
    inline bignum operator+(const int a)
    {
        bignum ret ;
        memcpy(ret.num,num,sizeof (bignum_t));
        ret.sgn=sgn ;
        ret+=a ;
        return ret ;
    }
    inline bignum&operator-=(const bignum&a)
    {
        if(sgn*a.sgn<0)add(num,a.num);
        else if
        (sgn&&a.sgn)
        {
            int ret=comp(num,a.num);
            if(ret>0)sub(num,a.num);
            else if(ret<0)
            {
                bignum_t t ;
                memcpy(t,num,sizeof(bignum_t));
                memcpy(num,a.num,sizeof(bignum_t));
                sub(num,t);
                sgn=-sgn ;
            }
            else memset(num,0,sizeof(bignum_t)),num[0]=1,sgn=0 ;
        }
        else if(!sgn)add (num,a.num),sgn=-a.sgn ;
        return*this ;
    }
    inline bignum&operator-=(const int a)
    {
        if(sgn*a<0)add(num,ABS(a));
        else if(sgn&&a)
        {
            int  ret=comp(num,ABS(a));
            if(ret>0)sub(num,ABS(a));
            else if(ret<0)
            {
                bignum_t t ;
                memcpy(t,num,sizeof(bignum_t));
                memset(num,0,sizeof(bignum_t));
                num[0]=1 ;
                add(num,ABS(a));
                sub(num,t);
                sgn=-sgn ;
            }
            else memset(num,0,sizeof(bignum_t)),num[0]=1,sgn=0 ;
        }
        else if
            (!sgn)sgn=-SGN(a),add(num,ABS(a));
        return*this ;
    }
    inline bignum operator-(const bignum&a)
    {
        bignum ret ;
        memcpy(ret.num,num,sizeof(bignum_t));
        ret.sgn=sgn ;
        ret-=a ;
        return ret ;
    }
    inline bignum operator-(const int a)
    {
        bignum ret ;
        memcpy(ret.num,num,sizeof(bignum_t));
        ret.sgn=sgn ;
        ret-=a ;
        return ret ;
    }
    inline bignum&operator*=(const bignum&a)
    {
        bignum_t t ;
        mul(t,num,a.num);
        memcpy(num,t,sizeof(bignum_t));
        sgn*=a.sgn ;
        return*this ;
    }
    inline bignum&operator*=(const int a)
    {
        mul(num,ABS(a));
        sgn*=SGN(a);
        return*this ;
    }
    inline bignum operator*(const bignum&a)
    {
        bignum ret ;
        mul(ret.num,num,a.num);
        ret.sgn=sgn*a.sgn ;
        return ret ;
    }
    inline bignum operator*(const int a)
    {
        bignum ret ;
        memcpy(ret.num,num,sizeof (bignum_t));
        mul(ret.num,ABS(a));
        ret.sgn=sgn*SGN(a);
        return ret ;
    }
    inline bignum&operator/=(const bignum&a)
    {
        bignum_t t ;
        div(t,num,a.num);
        memcpy (num,t,sizeof(bignum_t));
        sgn=(num[0]==1&&!num[1])?0:sgn*a.sgn ;
        return*this ;
    }
    inline bignum&operator/=(const int a)
    {
        int t ;
        div(num,ABS(a),t);
        sgn=(num[0]==1&&!num [1])?0:sgn*SGN(a);
        return*this ;
    }
    inline bignum operator/(const bignum&a)
    {
        bignum ret ;
        bignum_t t ;
        memcpy(t,num,sizeof(bignum_t));
        div(ret.num,t,a.num);
        ret.sgn=(ret.num[0]==1&&!ret.num[1])?0:sgn*a.sgn ;
        return ret ;
    }
    inline bignum operator/(const int a)
    {
        bignum ret ;
        int t ;
        memcpy(ret.num,num,sizeof(bignum_t));
        div(ret.num,ABS(a),t);
        ret.sgn=(ret.num[0]==1&&!ret.num[1])?0:sgn*SGN(a);
        return ret ;
    }
    inline bignum&operator%=(const bignum&a)
    {
        bignum_t t ;
        div(t,num,a.num);
        if(num[0]==1&&!num[1])sgn=0 ;
        return*this ;
    }
    inline int operator%=(const int a)
    {
        int t ;
        div(num,ABS(a),t);
        memset(num,0,sizeof (bignum_t));
        num[0]=1 ;
        add(num,t);
        return t ;
    }
    inline bignum operator%(const bignum&a)
    {
        bignum ret ;
        bignum_t t ;
        memcpy(ret.num,num,sizeof(bignum_t));
        div(t,ret.num,a.num);
        ret.sgn=(ret.num[0]==1&&!ret.num [1])?0:sgn ;
        return ret ;
    }
    inline int operator%(const int a)
    {
        bignum ret ;
        int t ;
        memcpy(ret.num,num,sizeof(bignum_t));
        div(ret.num,ABS(a),t);
        memset(ret.num,0,sizeof(bignum_t));
        ret.num[0]=1 ;
        add(ret.num,t);
        return t ;
    }
    inline bignum&operator++()
    {
        *this+=1 ;
        return*this ;
    }
    inline bignum&operator--()
    {
        *this-=1 ;
        return*this ;
    }
    ;
    inline int operator>(const bignum&a)
    {
        return sgn>0?(a.sgn>0?comp(num,a.num)>0:1):(sgn<0?(a.sgn<0?comp(num,a.num)<0:0):a.sgn<0);
    }
    inline int operator>(const int a)
    {
        return sgn>0?(a>0?comp(num,a)>0:1):(sgn<0?(a<0?comp(num,-a)<0:0):a<0);
    }
    inline int operator>=(const bignum&a)
    {
        return sgn>0?(a.sgn>0?comp(num,a.num)>=0:1):(sgn<0?(a.sgn<0?comp(num,a.num)<=0:0):a.sgn<=0);
    }
    inline int operator>=(const int a)
    {
        return sgn>0?(a>0?comp(num,a)>=0:1):(sgn<0?(a<0?comp(num,-a)<=0:0):a<=0);
    }
    inline int operator<(const bignum&a)
    {
        return sgn<0?(a.sgn<0?comp(num,a.num)>0:1):(sgn>0?(a.sgn>0?comp(num,a.num)<0:0):a.sgn>0);
    }
    inline int operator<(const int a)
    {
        return sgn<0?(a<0?comp(num,-a)>0:1):(sgn>0?(a>0?comp(num,a)<0:0):a>0);
    }
    inline int operator<=(const bignum&a)
    {
        return sgn<0?(a.sgn<0?comp(num,a.num)>=0:1):(sgn>0?(a.sgn>0?comp(num,a.num)<=0:0):a.sgn>=0);
    }
    inline int operator<=(const int a)
    {
        return sgn<0?(a<0?comp(num,-a)>=0:1):
        (sgn>0?(a>0?comp(num,a)<=0:0):a>=0);
    }
    inline int operator==(const bignum&a)
    {
        return(sgn==a.sgn)?!comp(num,a.num):0 ;
    }
    inline int operator==(const int a)
    {
        return(sgn*a>=0)?!comp(num,ABS(a)):0 ;
    }
    inline int operator!=(const bignum&a)
    {
        return(sgn==a.sgn)?comp(num,a.num):1 ;
    }
    inline int operator!=(const int a)
    {
        return(sgn*a>=0)?comp(num,ABS(a)):1 ;
    }
    inline int operator[](const int a)
    {
        return digit(num,a);
    }
    friend inline istream&operator>>(istream&is,bignum&a)
    {
        read(a.num,a.sgn,is);
        return  is ;
    }
    friend inline ostream&operator<<(ostream&os,const bignum&a)
    {
        if(a.sgn<0)
            os<<‘-‘ ;
        write(a.num,os);
        return os ;
    }
    friend inline bignum sqrt(const bignum&a)
    {
        bignum ret ;
        bignum_t t ;
        memcpy(t,a.num,sizeof(bignum_t));
        sqrt(ret.num,t);
        ret.sgn=ret.num[0]!=1||ret.num[1];
        return ret ;
    }
    friend inline bignum sqrt(const bignum&a,bignum&b)
    {
        bignum ret ;
        memcpy(b.num,a.num,sizeof(bignum_t));
        sqrt(ret.num,b.num);
        ret.sgn=ret.num[0]!=1||ret.num[1];
        b.sgn=b.num[0]!=1||ret.num[1];
        return ret ;
    }
    inline int length()
    {
        return :: length(num);
    }
    inline int zeronum()
    {
        return :: zeronum(num);
    }
    inline bignum C(const int m,const int n)
    {
        combination(num,m,n);
        sgn=1 ;
        return*this ;
    }
    inline bignum P(const int m,const int n)
    {
        permutation(num,m,n);
        sgn=1 ;
        return*this ;
    }
};
int main()
{
    bignum n, k;
    while(cin>>n)
    {
        if(n%2==0)
        {
            n = n/2;
            if(n%2==0)
                n = n-1;
            else
                n = n-2;
            cout<<n<<endl;
        }
        else
            cout<<n/2<<endl;
    }
    return 0 ;
}