High Precision computing 2

High Precision Division:

 

# Include <stdio. h> <br/> # include <string. h> </P> <p> # define max_len 200 </P> <p> char szline1 [max_len + 10]; <br/> char szline2 [max_len + 10]; <br/> int an1 [max_len + 10]; // divisor, an1 [0] corresponds to a single digit <br/> int an2 [max_len + 10]; // divisor, an2 [0] corresponds to a single digit <br/> int aresult [max_len + 10]; // storage provider, aresult [0] corresponds to a single digit </P> <p> // The Big integer P1 with the length of plen1 sends a big integer P2 with the length of nlen2 <br/> int substract (int * p1, int * P2, int nlen1, int nlen2) <br/>{< br/> int I; </ P> <p> // determine whether P1 is larger than P2. If not, return-1 <br/> If (nlen1 <nlen2) <br/> return-1; <br/> If (nlen1 = nlen2) <br/> for (I = nlen1-1; I> = 0; I --) <br/> If (P1 [I]> P2 [I]) <br/> break; <br/> else if (P1 [I] <P2 [I]) <br/> return-1; <br/> for (I = 0; I <nlen1; I ++) {<br/> // call this function to ensure that when I> = nlen2, P2 [I] = 0 <br/> P1 [I]-= P2 [I]; <br/> If (P1 [I] <0) {<br/> P1 [I] + = 10; <br/> P1 [I + 1] --; <br/>}< br/> for (I = nlen1- 1; I> = 0; I --) <br/> If (P1 [I]) // find the first of the highest bits, not 0 <br/> return I + 1; <br/> return 0; // All values are 0, indicating that the values are equal again <br/>}</P> <p> int main (void) <br/>{< br/> int t, n; </P> <p> scanf ("% d", & N ); <br/> for (t = 0; t <n; t ++) {<br/> scanf ("% s", szline1 ); <br/> scanf ("% s", szline2); <br/> int I, j; <br/> int nlen1 = strlen (szline1 ); <br/> memset (an1, 0, sizeof (an1); <br/> memset (an2, 0, sizeof (an2); <br/> memset (aresult, 0, sizeof (Aresult); <br/> for (j = 0, I = nlen1-1; I> = 0; I --) <br/> an1 [J ++] = szline1 [I]-'0'; <br/> int nlen2 = strlen (szline2 ); <br/> for (j = 0, I = nlen2-1; I> = 0; I --) <br/> an2 [J ++] = szline2 [I]-'0'; <br/> If (nlen1 <nlen1) {<br/> printf ("0/N"); <br/> continue; <br/>}</P> <p> int ntimes = nlen1-nlen2; <br/> If (ntimes> 0) {<br/> for (I = nlen1-1; I >= ntimes; I --) <br/> an2 [I] = an2 [I-ntimes ]; // Move toward the high position <br/> for (; I> = 0; I --) // fill 0 in the low position <br/> an2 [I] = 0; <br/> nlen2 = nlen1; <br/>}< br/> for (j = 0; j <= ntimes; j ++) {<br/> int ntmp; <br/> // until it is enough <br/> // first subtract several an2 * (10 to the power of ntimes), <br/> // not enough rebellion, then subtract several an2 * (10 to the power of the nTimes-1 ),... <br/> while (ntmp = substract (an1, an2 + J, nlen1, nlen2-j) >=0) {<br/> nlen1 = ntmp; <br/> aresult [ntimes-J] ++; // subtract each success, add the corresponding bid of the operator to 1 <br/>}< br/> // The output result below. Skip the high 0 <br/> for (I = max_len; I >=0 &&! Aresult [I]; I --) <br/>; <br/> if (I> = 0) <br/> for (; I> = 0; I --) <br/> printf ("% d", aresult [I]); <br/> else <br/> printf ("0 "); <br/> putchar ('/N'); </P> <p >}</P> <p> return 0; <br/>}

High-Precision multiplication:

 # Include <stdio. h> <br/> # include <string. h> </P> <p> # define max_len 200 </P> <p> int main (void) <br/>{< br/> int I, j, len1, len2; <br/> int A [max_len + 10], B [max_len + 10], C [max_len * 2 + 10]; <br/> char str1 [max_len + 10], str2 [max_len + 10]; </P> <p> // initialization <br/> for (I = 0; I <max_len + 10; I ++) <br/> A [I] = B [I] = 0; <br/> for (I = 0; I <max_len * 2 + 10; I ++) <br/> C [I] = 0; </P> <p> // enter two multiplier values. <br/> gets (St R1); <br/> gets (str2); </P> <p> // stores the number in an integer array in reverse order. <br/> len1 = strlen (str1 ); <br/> for (j = 0, I = len1-1; I> = 0; I --) <br/> A [J ++] = str1 [I]-'0'; <br/> len2 = strlen (str2 ); <br/> for (j = 0, I = len2-1; I> = 0; I --) <br/> B [J ++] = str2 [I]-'0'; </P> <p> for (I = 0; I <len2; I ++) // The second number multiplied by the first number. Each time one digit <br/> for (j = 0; j <len1; j ++) <br/> C [I + J] = B [I] * A [J]; // first multiply, followed by a unified carry <br/> for (I = 0; I <max_len * 2; I ++) <br/>{< br/> If (C [I]> = 10) {<br/> C [I + 1] + = C [I]/10; <br/> C [I] % = 10; <br/>}</P> <p> // output the multiplication result <br/> for (I = max_len * 2 ;! C [I] & I> = 0; I --) // crossing the high 0 <br/>; <br/> if (I> = 0) <br/> for (; I >= 0; I --) <br/> printf ("% d", C [I]); <br/> else <br/> printf ("0"); </P> <p> return 0; <br/>}

 

The last five hundred digits and total digits of msen

(You do not need to determine whether it is a prime number)

Formula: 2 ^ P-1

# Include <stdio. h> <br/> # include <memory. h> <br/> # include <math. h> </P> <p> # define Len 125 // each element contains four digits, the array requires a maximum of 125 elements </P> <p> // calculate the high-precision multiplication a * B, the result contains no 500 bits in a. <br/> void multiply (int * a, int * B) <br/>{< br/> int I, J; <br/> int ncarry; // memory carry <br/> int ntmp; <br/> int C [Len]; // store the last 500 bits of the result </P> <p> memset (C, 0, sizeof (INT) * Len); <br/> for (I = 0; I <Len; I ++) {<br/> ncarry = 0; <br/> for (j = 0; j <len-I; j ++) {<br/> Ntmp = C [I + J] + A [J] * B [I] + ncarry; <br/> C [I + J] = ntmp % 10000; <br/> ncarry = ntmp/10000; <br/>}< br/> memcpy (A, C, Len * sizeof (INT )); <br/>}</P> <p> int main (void) <br/>{< br/> int I, P; <br/> int anpow [Len]; // store the increasing power of 2 <br/> int aresult [Len]; // store the last 500 bits of the final result </P> <p> scanf ("% d", & P); <br/> printf ("% d/N ", (INT) (p * log10 (2.0) + 1); <br/> // initialize the power of 2 2 2 ^ (2 ^ 0) (A ^ B indicates the power B of A) <br/> // the final result is initialized to 1. <Br/> anpow [0] = 2; <br/> aresult [0] = 1; <br/> for (I = 1; I <Len; I ++) {<br/> anpow [I] = 0; <br/> aresult [I] = 0; <br/>}< br/> while (P> 0) {// calculate the power of 2 to the power of P <br/> // if p = 0, the valid bits in are used, if <br/> If (P & 1) // you do not need to calculate if the percentile bit in P is 1 <br/> multiply (aresult, anpow ); <br/> P> = 1; <br/> multiply (anpow, anpow); <br/>}< br/> aresult [0] --; // 2 ^ p-1 <br/> // output result <br/> for (I = len-1; I> = 0; I --) {<br/> if (I % 25 = 12) <Br/> printf ("% 02d/n % 02d", aresult [I]/100, aresult [I] % 100 ); <br/> else {<br/> printf ("% 04d", aresult [I]); <br/> If (! (I % 25) <br/> putchar ('/N'); <br/>}</P> <p> return 0; <br/>}