Acdream Chinese Girls' Amusement
A - Chinese Girls' Amusement
Problem Description
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
Output
Sample Input
7
6
Sample Output
3
1
Hint
/*
* this code is made by 987690183
* Problem: 1210
* Verdict: Accepted
* Submission Date: 2014-10-14 13:59:15
* Time: 0MS
* Memory: 1680KB
*/
#include<iostream>
#include<cstring>
#include<iomanip>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
using namespace std; #define MAXN 9999
#define MAXSIZE 1000
#define DLEN 4 class BigNum
{
private:
int a[MAXSIZE]; //可以控制大数的位数
int len; //大数长度
public:
BigNum(){ len = ;memset(a,,sizeof(a)); } //构造函数
BigNum(const int); //将一个int类型的变量转化为大数
BigNum(const char*); //将一个字符串类型的变量转化为大数
BigNum(const BigNum &); //拷贝构造函数
BigNum &operator=(const BigNum &); //重载赋值运算符,大数之间进行赋值运算 friend istream& operator>>(istream&, BigNum&); //重载输入运算符
friend ostream& operator<<(ostream&, BigNum&); //重载输出运算符 BigNum operator+(const BigNum &) const; //重载加法运算符,两个大数之间的相加运算
BigNum operator-(const BigNum &) const; //重载减法运算符,两个大数之间的相减运算
BigNum operator*(const BigNum &) const; //重载乘法运算符,两个大数之间的相乘运算
BigNum operator/(const int &) const; //重载除法运算符,大数对一个整数进行相除运算 BigNum operator^(const int &) const; //大数的n次方运算
int operator%(const int &) const; //大数对一个int类型的变量进行取模运算
bool operator>(const BigNum & T)const; //大数和另一个大数的大小比较
bool operator>(const int & t)const; //大数和一个int类型的变量的大小比较 void print(); //输出大数
};
BigNum::BigNum(const int b) //将一个int类型的变量转化为大数
{
int c,d = b;
len = ;
memset(a,,sizeof(a));
while(d > MAXN)
{
c = d - (d / (MAXN + )) * (MAXN + );
d = d / (MAXN + );
a[len++] = c;
}
a[len++] = d;
}
BigNum::BigNum(const char*s) //将一个字符串类型的变量转化为大数
{
int t,k,index,l,i;
memset(a,,sizeof(a));
l=strlen(s);
len=l/DLEN;
if(l%DLEN)
len++;
index=;
for(i=l-;i>=;i-=DLEN)
{
t=;
k=i-DLEN+;
if(k<)
k=;
for(int j=k;j<=i;j++)
t=t*+s[j]-'';
a[index++]=t;
}
}
BigNum::BigNum(const BigNum & T) : len(T.len) //拷贝构造函数
{
int i;
memset(a,,sizeof(a));
for(i = ; i < len ; i++)
a[i] = T.a[i];
}
BigNum & BigNum::operator=(const BigNum & n) //重载赋值运算符,大数之间进行赋值运算
{
int i;
len = n.len;
memset(a,,sizeof(a));
for(i = ; i < len ; i++)
a[i] = n.a[i];
return *this;
}
istream& operator>>(istream & in, BigNum & b) //重载输入运算符
{
char ch[MAXSIZE*];
int i = -;
in>>ch;
int l=strlen(ch);
int count=,sum=;
for(i=l-;i>=;)
{
sum = ;
int t=;
for(int j=;j<&&i>=;j++,i--,t*=)
{
sum+=(ch[i]-'')*t;
}
b.a[count]=sum;
count++;
}
b.len =count++;
return in; }
/*ostream& operator<<(ostream& out, BigNum& b) //重载输出运算符
{
int i;
cout << b.a[b.len - 1];
for(i = b.len - 2 ; i >= 0 ; i--)
{
cout.width(DLEN);
cout.fill('0');
cout << b.a[i];
}
return out;
}*/ BigNum BigNum::operator+(const BigNum & T) const //两个大数之间的相加运算
{
BigNum t(*this);
int i,big; //位数
big = T.len > len ? T.len : len;
for(i = ; i < big ; i++)
{
t.a[i] +=T.a[i];
if(t.a[i] > MAXN)
{
t.a[i + ]++;
t.a[i] -=MAXN+;
}
}
if(t.a[big] != )
t.len = big + ;
else
t.len = big;
return t;
}
BigNum BigNum::operator-(const BigNum & T) const //两个大数之间的相减运算
{
int i,j,big;
bool flag;
BigNum t1,t2;
if(*this>T)
{
t1=*this;
t2=T;
flag=;
}
else
{
t1=T;
t2=*this;
flag=;
}
big=t1.len;
for(i = ; i < big ; i++)
{
if(t1.a[i] < t2.a[i])
{
j = i + ;
while(t1.a[j] == )
j++;
t1.a[j--]--;
while(j > i)
t1.a[j--] += MAXN;
t1.a[i] += MAXN + - t2.a[i];
}
else
t1.a[i] -= t2.a[i];
}
t1.len = big;
while(t1.a[len - ] == && t1.len > )
{
t1.len--;
big--;
}
if(flag)
t1.a[big-]=-t1.a[big-];
return t1;
} BigNum BigNum::operator*(const BigNum & T) const //两个大数之间的相乘运算
{
BigNum ret;
int i,j,up;
int temp,temp1;
for(i = ; i < len ; i++)
{
up = ;
for(j = ; j < T.len ; j++)
{
temp = a[i] * T.a[j] + ret.a[i + j] + up;
if(temp > MAXN)
{
temp1 = temp - temp / (MAXN + ) * (MAXN + );
up = temp / (MAXN + );
ret.a[i + j] = temp1;
}
else
{
up = ;
ret.a[i + j] = temp;
}
}
if(up != )
ret.a[i + j] = up;
}
ret.len = i + j;
while(ret.a[ret.len - ] == && ret.len > )
ret.len--;
return ret;
}
BigNum BigNum::operator/(const int & b) const //大数对一个整数进行相除运算
{
BigNum ret;
int i,down = ;
for(i = len - ; i >= ; i--)
{
ret.a[i] = (a[i] + down * (MAXN + )) / b;
down = a[i] + down * (MAXN + ) - ret.a[i] * b;
}
ret.len = len;
while(ret.a[ret.len - ] == && ret.len > )
ret.len--;
return ret;
}
int BigNum::operator %(const int & b) const //大数对一个int类型的变量进行取模运算
{
int i,d=;
for (i = len-; i>=; i--)
{
d = ((d * (MAXN+))% b + a[i])% b;
}
return d;
}
BigNum BigNum::operator^(const int & n) const //大数的n次方运算
{
BigNum t,ret();
int i;
if(n<)
exit(-);
if(n==)
return ;
if(n==)
return *this;
int m=n;
while(m>)
{
t=*this;
for( i=;i<<<=m;i<<=)
{
t=t*t;
}
m-=i;
ret=ret*t;
if(m==)
ret=ret*(*this);
}
return ret;
}
bool BigNum::operator>(const BigNum & T) const //大数和另一个大数的大小比较
{
int ln;
if(len > T.len)
return true;
else if(len == T.len)
{
ln = len - ;
while(a[ln] == T.a[ln] && ln >= )
ln--;
if(ln >= && a[ln] > T.a[ln])
return true;
else
return false;
}
else
return false;
}
bool BigNum::operator >(const int & t) const //大数和一个int类型的变量的大小比较
{
BigNum b(t);
return *this>b;
} void BigNum::print() //输出大数
{
int i;
//cout << a[len - 1];
printf("%d",a[len-]);
for(i = len - ; i >= ; i--)
{
/*cout.width(DLEN);
cout.fill('0');
cout << a[i];*/
printf("%04d",a[i]);
}
//cout << endl;
printf("\n");
}
int main()
{
char zero[]={""};
char one[]={""};
char two[]={""};
char hxl[];
BigNum z,ZERO(zero),ONE(one),TWO(two);
while(gets(hxl)>)
{
BigNum x(hxl);
z=x%;
if(z>)/**奇数**/
{
z=(x-ONE)/;
}
else /**偶数**/
{
x=x/;
z=x%;
if(z>)/**奇数**/
{
z=x-TWO;
}
else /**偶数**/
{
z=x-ONE;
}
}
z.print();
}
return ;
}
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