[高精度][codevs 1145]Hanoi双塔问题

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#define maxn 2000
#define base 10000
using namespace std;
int n;
struct Bign
{
int c[maxn],len,sign;
//初始化
Bign(){memset(c,0,sizeof(c)),len = 1,sign = 0;}
//高位清零
void Zero()
{
while(len > 1 && c[len] == 0)len--;
if(len == 1 && c[len] == 0)sign = 0;
}
//压位读入
void Write(char *s)
{
int k = 1,l = strlen(s);
for(int i = l - 1;i >= 0;i--)
{
c[len] += (s[i] - '0') * k;
k *= 10;
if(k == base)
{
k = 1;
len++;
}
}
}
void Read()
{
char s[maxn] = {0};
scanf("%s",s);
Write(s);
}
//输出
void Print()
{
if(sign)printf("-");
printf("%d",c[len]);
for(int i = len - 1;i >= 1;i--)printf("%04d",c[i]);
printf("\n");
}
//重载 = 运算符，将低精赋值给高精
Bign operator = (int a)
{
char s[100];
sprintf(s,"%d",a);
Write(s);
return *this;//this只能用于成员函数，表示当前对象的地址
}
//重载 < 运算符
bool operator < (const Bign &a)const
{
if(len != a.len)return len < a.len;
for(int i = len;i >= 1;i--)
{
if(c[i] != a.c[i])return c[i] < a.c[i];
}
return 0;
}
bool operator > (const Bign &a)const
{
return a < *this;
}
bool operator <= (const Bign &a)const
{
return !(a < *this);
}
bool operator >= (const Bign &a)const
{
return !(*this < a);
}
bool operator != (const Bign &a)const
{
return a < *this || *this < a;
}
bool operator == (const Bign &a)const
{
return !(a < *this) && !(*this < a);
}
bool operator == (const int &a)const
{
Bign b;b = a;
return *this == b;
}

//重载 + 运算符
Bign operator + (const Bign &a)
{
Bign r;
r.len = max(len,a.len) + 1;
for(int i = 1;i <= r.len;i++)
{
r.c[i] += c[i] + a.c[i];
r.c[i + 1] += r.c[i] / base;
r.c[i] %= base;
}
r.Zero();
return r;
}
Bign operator + (const int &a)
{
Bign b;b = a;
return *this + b;
}
//重载 - 运算符
Bign operator - (const Bign &a)
{
Bign b,c;// b - c
b = *this;
c = a;
if(c > b)
{
swap(b,c);
b.sign = 1;
}
for(int i = 1;i <= b.len;i++)
{
b.c[i] -= c.c[i];
if(b.c[i] < 0)
{
b.c[i] += base;
b.c[i + 1]--;
}
}
b.Zero();
return b;
}
Bign operator - (const int &a)
{
Bign b;b = a;
return *this - b;
}
//重载 * 运算符
Bign operator * (const Bign &a)
{
Bign r;
r.len = len + a.len + 2;
for(int i = 1;i <= len;i++)
{
for(int j = 1;j <= a.len;j++)
{
r.c[i + j - 1] += c[i] * a.c[j];
}
}
for(int i = 1;i <= r.len;i++)
{
r.c[i + 1] += r.c[i] / base;
r.c[i] %= base;
}
r.Zero();
return r;
}
Bign operator * (const int &a)
{
Bign b;b = a;
return *this * b;
}
//重载 / 运算符
Bign operator / (const Bign &b)
{
Bign r,t,a;
a = b;
//if(a == 0)return r;
r.len = len;
for(int i = len;i >= 1;i--)
{
t = t * base + c[i];
int div,ll = 0,rr = base;
while(ll <= rr)
{
int mid = (ll + rr) / 2;
Bign k = a * mid;
if(k > t)rr = mid - 1;
else
{
ll = mid + 1;
div = mid;
}
}
r.c[i] = div;
t = t - a * div;
}
r.Zero();
return r;
}
Bign operator / (const int &a)
{
Bign b;b = a;
return *this / b;
}
//重载 % 运算符
Bign operator % (const Bign &a)
{
return *this - *this / a * a;
}
Bign operator % (const int &a)
{
Bign b;b = a;
return *this % b;
}
};

int main()
{
scanf("%d",&n);
Bign ans;
ans =1;
for(int i=1;i<=n+1;i++)
ans=ans+ans;
ans=ans-2;
ans.Print();
return 0;
}