HDU 1134 modulo of large numbers use a large number Template

# Include <iostream> # include <string> # include <iomanip> # include <algorithm> using namespace STD; char STR [1010]; int modd; # define maxn 9999 # define maxsize 10 # define dlen 4 class bignum {PRIVATE: int A [1010]; // The number of digits that can be controlled for a large number int Len; // the length of a large number is public: bignum () {Len = 1; memset (A, 0, sizeof (a);} // constructor bignum (const INT ); // convert a variable of the int type to the bignum (const char *); // convert a variable of the string type to the bignum (const bignum &); // copy the constructor bignu M & operator = (const bignum &); // reload the value assignment operator to assign values between large numbers: Friend istream & operator> (istream &, bignum &); // reload the input operator friend ostream & operator <(ostream &, bignum &); // reload the output operator bignum operator + (const bignum &) const; // reload the addition operator, bignum operator-(const bignum &) const; // overload subtraction operator, subtraction between two large numbers bignum operator * (const bignum &) const; // reload the multiplication operator. The multiplication operation between two large numbers is bignum operator/(const Int &) const; // reload the Division operator. Integer Division bignum operator ^ (const Int &) const; // Npower operation of large numbers int operator % (const Int &) const; // perform the modulo operation bool operator> (const bignum & T) const on an int type variable in a large number; // compare the size of a large number with that of another large number bool operator> (const Int & T) const; // compare the size of a large number with that of an int Type Variable Void print (); // output large number}; bignum: bignum (const int B) // convert an int type variable to a large number {int C, D = B; Len = 0; memset (A, 0, sizeof (a); While (D> maxn) {c = D-(D/(maxn + 1) * (maxn + 1 ); D = D/(MA Xn + 1); A [Len ++] = C;} A [Len ++] = D;} bignum: bignum (const char * s) // convert a variable of the string type to a large number {int T, K, index, l, I; memset (A, 0, sizeof ()); L = strlen (s); Len = L/dlen; If (L % dlen) Len ++; Index = 0; for (I = L-1 I> = 0; i-= dlen) {T = 0; k = I-dlen + 1; if (k <0) k = 0; For (Int J = K; j <= I; j ++) t = T * 10 + s [J]-'0'; A [index ++] = T ;}} bignum: bignum (const bignum & T): Len (T. len) // copy the constructor {int I; memset (A, 0, sizeof (a); for (I = 0; I <Len; I ++) A [I] = T. A [I];} Bignum & bignum: Operator = (const bignum & N) // reload the value assignment operator and assign values between large numbers {int I; Len = n. len; memset (A, 0, sizeof (a); for (I = 0; I <Len; I ++) A [I] = n. A [I]; return * This;} istream & operator> (istream & in, bignum & B) // reload the input operator {char ch [maxsize * 4]; int I =-1; in> CH; int L = strlen (CH); int COUNT = 0, sum = 0; for (I = L-1; I> = 0 ;) {sum = 0; int T = 1; for (Int J = 0; j <4 & I> = 0; j ++, I --, T * = 10) {sum + = (CH [I]-'0') * t;} B. A [count] = sum; Count ++;} B. len = count ++; Return in;} ostream & operator <(ostream & out, bignum & B) // reload the output operator {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: Operator + (const bignum & T) const // The addition operation between two large numbers {bignum T (* This ); int I, big; // The number of digits big = T. len> Len? T. len: Len; for (I = 0; I <big; I ++) {T. A [I] + = T. A [I]; If (T. A [I]> maxn) {T. A [I + 1] ++; T. A [I]-= maxn + 1 ;}} if (T. A [Big]! = 0) T. len = big + 1; elset. len = big; return t;} bignum: Operator-(const bignum & T) const // subtraction between two large numbers {int I, j, big; bool flag; bignum T1, T2; If (* This> T) {T1 = * This; t2 = T; flag = 0;} else {T1 = T; t2 = * this; flag = 1 ;}big = t1.len; for (I = 0; I <big; I ++) {If (t1.a [I] <t2.a [I]) {J = I + 1; while (t1.a [J] = 0) J ++; t1.a [j --] --; while (j> I) t1.a [j --] + = maxn; t1.a [I] + = maxn + 1-t2.a [I];} elset1.a [I]-= t2.a [I];} t1.len = big; while (t1.a [Len-1] = 0 & t1.len> 1) {t1.len --; big --;} If (FLAG) t1.a [Big-1] = 0-t1.a [Big-1]; return T1;} bignum: Operator * (const bignum & T) const // multiplication between two large numbers {bignum ret; int I, j, up; int temp, temp1; for (I = 0; I <Len; I ++) {up = 0; For (j = 0; j <t. len; j ++) {temp = A [I] * t. A [J] + ret. A [I + J] + up; If (temp> maxn) {temp1 = temp-temp/(maxn + 1) * (maxn + 1); up = Temp/(maxn + 1); ret. A [I + J] = temp1;} else {up = 0; ret. A [I + J] = temp ;}} if (up! = 0) ret. A [I + J] = up;} ret. len = I + J; while (Ret. A [ret. len-1] = 0 & ret. len> 1) ret. len --; return ret;} bignum: Operator/(const Int & B) const // perform the division operation on an integer {bignum ret; int I, down = 0; for (I = len-1; I> = 0; I --) {ret. A [I] = (a [I] + down * (maxn + 1)/B; down = A [I] + down * (maxn + 1)-ret. A [I] * B;} ret. len = Len; while (Ret. A [ret. len-1] = 0 & ret. len> 1) ret. len --; Return ret;} int bignum: Operator % (const Int & B) const // a large number of Modulo operations on an int type variable {int I, D = 0; for (I = len-1; I> = 0; I --) {d = (D * (maxn + 1) % B + A [I]) % B ;} return D;} bignum: operator ^ (const Int & N) const // Npower operation of large numbers {bignum T, RET (1); int I; if (n <0) Exit (-1); If (n = 0) return 1; if (n = 1) return * This; int M = N; while (M> 1) {T = * this; for (I = 1; I <1 <= m; I <= 1) {T = T * t ;} m-= I; ret = RET * t; If (M = 1) ret = RET * (* This);} return RET;} bool bignum: Operator> (const bignum & T) const // compare the size of a large number with that of another one {int ln; If (LEN> T. len) return true; else if (LEN = T. len) {Ln = len-1; while (A [Ln] = T. A [Ln] & ln> = 0) ln --; If (LN> = 0 & A [Ln]> T. A [Ln]) return true; elsereturn false;} bool bignum: Operator> (const Int & T) const // compare the size of a large number and an int type variable {bignum B (t); return * This> B;} void bignum: Print () // output large number {int I; cout <A [Len-1]; for (I = len-2; I> = 0; I --) {cout. width (dlen); cout. fill ('0'); cout <A [I];} cout <Endl;} int main (void) {While (~ Scanf ("% S % d", STR, & modd) {bignum big (STR); cout <big % modd <Endl ;}return 0 ;}

Using a large number template, it is easy to be speechless... For more information about the template, see http://blog.csdn.net/vsooda/article/details/8543351.