[High accuracy] [Codevs 1145] Hanoi Twin Towers problem

#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;//Initialize Bign () { Memset (C,0,sizeof (c)), Len = 1,sign = 0;} High-clear 0 void Zero () {while (len > 1 && c[len] = = 0) len--;if (len = = 1 && C[len] = = 0) sign = 0;}//press bit read in 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);} output void Print () {if (sign) printf ("-");p rintf ("%d", C[len]), and for (int i = len-1;i >= 1;i--) printf ("%04d", C[i]);p rintf ( "\ n");} overloading = operator, low-fine assignment to high-precision bign operator = (int a) {char s[100];sprintf (s, "%d", a); Write (s); return *this;//this can only be used for member functions, representing the address of the current object}//overloading < operator 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];} Return0;} 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;}  Overloaded + operator 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;} Overloaded-operator 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;} Overloaded * operator 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;} Overloaded/Operator 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;} Overloaded% operator 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;}

[High accuracy] [Codevs 1145] Hanoi Twin Towers problem