[高精度][codevs 1145]Hanoi双塔问题

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#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<cmath>#define maxn 2000#define base 10000using 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 - cb = *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;}
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