High Accuracy-templates

Alas, I have been busy for a whole day, tired, tired, and tired ....

 

But for the sake of the AC business, it is still worth it ....

 

 

 

 

# Include <stdio. h> # include <string. h> # include <stdlib. h> # include <math. h> # include <assert. h> # include <ctype. h> # include <map> # include <string> # include <set> # include <bitset> # include <utility> # include <algorithm> # include <vector> # include <stack> # include <queue> # include <iostream> # include <fstream> # include <list> using namespace STD; const int maxl = 500; struct bignum {int num [maxl]; int Len ;}; // High Precision comparison> B return 1, A = B Return 0; A <B Return-1; int comp (bignum & A, bignum & B) {int I; If (A. Len! = B. Len) Return (A. Len> B. Len )? 1:-1; for (I = A. len-1; I> = 0; I --) if (A. Num [I]! = B. Num [I]) Return (A. Num [I]> B. Num [I])? 1:-1; return 0;} // Add bignum add (bignum & A, bignum & B) {bignum C; int I, Len; Len = (. len> B. len )? A. len: B. len; memset (C. num, 0, sizeof (C. num); for (I = 0; I <Len; I ++) {C. num [I] + = (. num [I] + B. num [I]); If (C. num [I]> = 10) {C. num [I + 1] ++; C. num [I]-= 10;} If (C. num [Len]) Len ++; C. len = Len; return C;} // high-precision subtraction to ensure that a> = bbignum sub (bignum & A, bignum & B) {bignum C; int I, Len; len = (. len> B. len )? A. len: B. len; memset (C. num, 0, sizeof (C. num); for (I = 0; I <Len; I ++) {C. num [I] + = (. num [I]-B. num [I]); If (C. num [I] <0) {C. num [I] + = 10; C. num [I + 1] --;} while (C. num [Len] = 0 & Len> 1) Len --; C. len = Len; return C;} // High Precision multiplied by low precision. When B is large, the int range may overflow, specific analysis // If B is large, consider B as a high-precision bignum mul1 (bignum & A, Int & B) {bignum C; int I, Len; Len =. len; memset (C. num, 0, sizeof (C. num); // multiply by 0 and return 0 if (B = 0) {C. len = 1; return C;} for (I = 0; I <Len; I ++) {C. num [I] + = (. num [I] * B); If (C. num [I]> = 10) {C. num [I + 1] = C. num [I]/10; C. num [I] % = 10;} while (C. num [Len]> 0) {C. num [Len + 1] = C. num [Len]/10; C. num [Len ++] % = 10;} C. len = Len; return C;} // High Precision multiplied by high precision. Be sure to carry data in time. Otherwise, overflow may occur, but this will increase the complexity of the algorithm, // if it is determined that no overflow will occur, you can change the while value to ifbignum mul2 (bignum & A, bignum & B) {int I, j, Len = 0; bignum C; memset (C. num, 0, Si Zeof (C. num); for (I = 0; I <. len; I ++) {for (j = 0; j <B. len; j ++) {C. num [I + J] + = (. num [I] * B. num [J]); If (C. num [I + J]> = 10) {C. num [I + J + 1] + = C. num [I + J]/10; C. num [I + J] % = 10 ;}} Len =. len + B. len-1; while (C. num [len-1] = 0 & Len> 1) Len --; If (C. num [Len]) Len ++; C. len = Len; return C;} // High Precision divided by low precision. The division result is C, and the remainder is fvoid div1 (bignum & A, Int & B, bignum & C, int & F) {int I, Len =. len; memset (C. num, 0, S Izeof (C. num); F = 0; for (I =. len-1; I> = 0; I --) {f = f * 10 +. num [I]; C. num [I] = f/B; F % = B;} while (LEN> 1 & C. num [len-1] = 0) Len --; C. len = Len;} // high precision * 10 void mul10 (bignum & A) {int I, Len =. len; for (I = Len; I> = 1; I --). num [I] =. num [I-1];. num [I] = 0; Len ++; // If a = 0 while (LEN> 1 &. num [len-1] = 0) Len --;} // High Precision divided by high precision, division result is C, remainder is fvoid div2 (bignum & A, bignum & B, bignum & C, bignum & F) {int I, Len =. len; memset (C. num, 0, sizeof (C. num); memset (F. num, 0, sizeof (F. num); F. len = 1; for (I = len-1; I> = 0; I --) {mul10 (f); // the remainder is multiplied by 10 F each time. num [0] =. num [I]; // then add the remainder to the next digit. // use subtraction to replace the Division while (COMP (F, B)> = 0) {f = sub (F, b); C. num [I] ++;} while (LEN> 1 & C. num [len-1] = 0) Len --; C. len = Len;} void print (bignum & A) // output large number {int I; for (I =. len-1; I> = 0; I --) printf ("% d",. num [I]); Pu TS ("") ;}// convert the string to a large number and the bignum tonum (char * s) {int I, j; bignum A;. len = strlen (s); for (I = 0, j =. len-1; s [I]! = '\ 0'; I ++, j --). num [I] = s [J]-'0'; return a;} void Init (bignum & A, char * s, Int & tag) // convert the string to a large number {int I = 0, j = strlen (s); If (s [0] = '-') {J --; I ++; tag * =-1;}. len = J; For (; s [I]! = '\ 0'; I ++, j --). num [J-1] = s [I]-'0';} int main () {bignum a, B; char S1 [100], S2 [100]; while (scanf ("% S % s", S1, S2 )! = EOF) {int tag = 1; Init (A, S1, tag); // convert a string to a large number of Init (B, S2, tag); A = add (, b); if (. len = 1 &. num [0] = 0) {puts ("0");} else {If (tag <0) putchar ('-'); print ();}} return 0 ;}