High-Precision computing template C Language

# Include <iostream> # include <string. h> # include <algorithm> # include <stdio. h> # define Maxx 100 using namespace STD; // high-precision addition: void add (char * S1, char * S2) {int n, m, I, J, A [Maxx + 10] = {0}, B [Maxx + 10] = {0}; int L1 = strlen (S1); int L2 = strlen (S2 ); for (I = l1-1, j = Maxx; I> = 0; I --, j --) A [J] = S1 [I]-'0 '; for (I = l2-1, j = Maxx; I> = 0; I --, j --) B [J] = S2 [I]-'0 '; for (I = Maxx; I> = 0; I --) {A [I] + = B [I]; if (a [I]> = 10) A [I-1] + = A [I]/10, a [I] % = 10;} int temp = 0; for (I = 0; I <= Maxx; I ++) {if (a [I]! = 0 & temp = 0) temp = 1; if (temp = 1) printf ("% d", a [I]);} puts ("");} // high-precision subtraction void sub (char * S1, char * S2) {int n, m, I, J, A [Maxx + 10] = {0}, B [Maxx + 10] = {0}, C [Maxx + 10] = {0 }; int L1 = strlen (S1); int L2 = strlen (S2); for (I = l1-1, j = Maxx; I> = 0; I --, j --) A [J] = S1 [I]-'0'; for (I = l2-1, j = Maxx; I> = 0; I --, j --) B [J] = S2 [I]-'0'; for (I = Maxx; I> = 0; I --) {c [I] + = A [I]-B [I]; If (C [I] <0) {C [I] + = 10; c [I-1] --;} int temp = 0; for (I = 0; I <= Maxx; I ++) {If (C [I]! = 0 & temp = 0) temp = 1; if (temp = 1) printf ("% d", C [I]);} puts ("");} // high-precision multiplication void MUL (char * S1, char * S2) {int n, m, I, j, i1, J1, A [Maxx + 10] = {0}, B [Maxx + 10] = {0}, C [2 * Maxx + 10] = {0 }; int L1 = strlen (S1); int L2 = strlen (S2); for (I = l1-1, j = 0; I> = 0; I --) A [J ++] = S1 [I]-'0'; for (I = l2-1, j = 0; I> = 0; I --) B [J ++] = S2 [I]-'0'; for (I = 0; I <L2; I ++) for (j = 0; j <L1; j ++) C [I + J] + = B [I] * A [J]; for (I = 0; I <Maxx * 2; I ++) if (C [I]> = 10) {C [I + 1] + = C [I]/10; C [I] % = 10;} int T = 0; For (I = Maxx * 2; I> = 0; I --) if (t) printf ("% d", C [I]); else if (C [I]) {printf ("% d", C [I]); t = 1 ;}if (! T) printf ("0"); puts ("");} // high-precision division # include <iostream> # include <string. h> # include <algorithm> # include <stdio. h> # define Maxx 100 using namespace STD; int an1 [Maxx + 10], an2 [Maxx + 10]; int ans [Maxx + 10];/* Sun function: A large integer P1 whose length is L1 minus a large integer P2 whose length is L2. The result is placed in P1. The returned value indicates the length of the result */INT sub (int * P1, int * P2, int L1, int l2) {int I; If (L1 <l2) Return-1; // check whether P1 is larger than P2. If not, returns-1 bool temp = false; If (L1 = l2) {for (I = l1-1; I> = 0; I --) {If (P1 [I]> P2 [I]) TEM P = true; else if (P1 [I] <P2 [I]) {If (! Temp) Return-1 ;}}for (I = 0; I <L1; I ++) {// do the subtraction P1 [I]-= P2 [I]; // The parameter required to call this function can ensure that when I> = len2, P2 [I] = 0 if (P1 [I] <0) {p1 [I] + = 10; P1 [I + 1] -- ;}} for (I = l1-1; I> = 0; I --) if (P1 [I]) return I + 1; return 0;} int main () {int I, j; char S1 [Maxx + 10], S2 [Maxx + 10]; scanf ("% S % s", S1, S2); int L1 = strlen (S1), L2 = strlen (S2); memset (an1, 0, sizeof (an1); memset (an2, 0, sizeof (an2); memset (ANS, 0, sizeof (ANS); for (I = l1-1, j = 0; i> = 0; I --) an1 [J ++] = S1 [I]-'0'; fo R (I = l2-1, j = 0; I> = 0; I --) an2 [J ++] = S2 [I]-'0'; If (L1 <l2) {printf ("0 \ n"); Return 0;} L1 = sub (an1, an2, L1, L2); If (L1 <0) {printf ("0 \ n"); Return 0;} else if (L1 = 0) {printf ("1 \ n"); Return 0 ;} ans [0] ++; int ntimes = L1-L2; If (ntimes <0) goto outputresult; else if (ntimes> 0) {for (I = l1-1; I> = 0; I --) {if (I> = ntimes) an2 [I] = an2 [I-ntimes]; elsean2 [I] = 0 ;}} L2 = L1; for (j = 0; j <= ntimes; j ++) {int ntmp; // until the limit is reached, first subtract several ans * (the ntimes power of 10) // not enough. Subtract again. Go to several ans * (10 ntimes power) while (ntmp = sub (an1, an2 + J, L1, l2-j)> = 0) {L1 = ntmp; ans [ntimes-J] ++ ;}} outputresult: // The following loop uniformly handles the carry problem for (I = 0; I <Maxx; I ++) {If (ANS [I] >=10) {ans [I + 1] + = ans [I]/10; ans [I] % = 10 ;}} bool T = false; for (I = Maxx; I> = 0; I --) if (t) printf ("% d", ANS [I]); else if (ANS [I]) {printf ("% d", ANS [I]); t = true;} If (! T) printf ("0 \ n"); printf ("\ n"); Return 0 ;}