Large number arithmetic and DH algorithm（大整数的运算与DH算法的实现）

按：这是信息安全原理课程的第二次作业，作业要求是Write a +-*/ algorithm for large integers. Implement the DH algorithm.

2014-4-27更新：过了DDL了，重新公开原先隐藏的代码（只是现在不能在代码片中打开，有人能告诉我要怎么做吗？）。补上了所有的图片。

Large integer algorithm

How to use

Open “\SEC2014-HW2-3130000064-林涛\large_integer_algorithm\large_integer_algorithm.exe”,input the value of A and B (it can be as long as 800 numbers, negative is alsoavailable). You can see the results of addition, subtraction, multiplication,division and
modulus.



Figure1 test for general largeintegers



Figure2 divide zero



Figure3 negative cases



Figure4 A is zero



Figure5 interesting case



Figure6 A has 300 numbers and Bhas 200 numbers

Algorithm

Use character array to store large number,and use the same way as primate school studens.

We can optimize the multiplicationalgorithm by calculating A*1, A*2 …, A*9 first, which don’t have to becalculated again and again.

Source code

large_integer_algorithm.c

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define SIZE 800
char answer[2 * SIZE];
char remainder[SIZE];
//a,b都为自然数，且a>=b
void sum1(char *a, char *b)
{
int carry = 0;
int pa, pb;
int temp;
int i;
int na = strlen(a);
int nb = strlen(b);

answer[na] = '\0';
pa = na - 1; pb = nb - 1;
while (pa >= 0) {
temp = (a[pa] - '0') + carry;
if (pb >= 0)temp += (b[pb] - '0');

if (temp >= 10) {
answer[pa] = temp - 10 + '0';
carry = 1;
}
else {
answer[pa] = temp + '0';
carry = 0;
}
pa--; pb--;
}
if (carry) {
for (i = na; i >= 0; i--)
answer[i + 1] = answer[i];
answer[0] = '1';
}
}

void sub1(char *a, char *b)
{
int borrow = 0;
int pa, pb, pc;
int temp;
int i;
int na = strlen(a);
int nb = strlen(b);

answer[na] = '\0';
pa = na - 1; pb = nb - 1;
while (pa >= 0) {
temp = (a[pa] - '0') - borrow;
if (pb >= 0)temp -= (b[pb] - '0');

if (temp < 0) {
answer[pa] = temp + 10 + '0';
borrow = 1;
}
else {
answer[pa] = temp + '0';
borrow = 0;
}
pa--; pb--;
}
pc = 0;
while (answer[pc] == '0')
pc++;
for (i = 0; i <= na - pc; i++)
answer[i] = answer[i + pc];
}
//a,b都为自然数
void sum(char *a, char *b)
{
int na = strlen(a);
int nb = strlen(b);

if (na >= nb)
sum1(a, b);
else
sum1(b, a);
}

void sub(char *a, char *b)
{
int na = strlen(a);
int nb = strlen(b);

if (strcmp(a, b) == 0)
strcpy(answer, "0");
else if (na > nb || na == nb && strcmp(a, b) > 0)
sub1(a, b);
else {
printf("-");
sub1(b, a);
}
}

//a,b都是正整数
void mul(char *a, char *b)
{
char muls[10][SIZE + 1];
char temp[2 * SIZE];
int i, j;
int na = strlen(a);
int nb = strlen(b);

strcpy(muls[0], "0");

strcpy(temp, "0");
for (i = 1; i <= 9; i++) {
sum(muls[i - 1], a);
strcpy(muls[i], answer);
}

strcpy(answer, "0");

for (i = nb - 1; i >= 0; i--) {
if (b[i] != '0') {
strcpy(temp, muls[b[i]-'0']);
//后面加（nb-i-1）个零
for (j = 0; j < nb - i - 1; j++)
strcat(temp, "0");
sum(answer, temp);
}
}

printf("%s\n", answer);

}
//a,b为自然数，若a>b返回1，不然0
int larger(char *a, char *b)
{
int na, nb;
na = strlen(a);
nb = strlen(b);
if (na > nb)
return 1;
else if (na == nb && strcmp(a, b) > 0)
return 1;
else
return 0;
}

void divide(char *a, char*b)
{
int na, nb; na = strlen(a); nb = strlen(b);
char quotient[SIZE];
char temp[2]; temp[1] = '\0';
int tryq;
int i;

char muls[10][SIZE + 1];
strcpy(muls[0], "0");
strcpy(temp, "0");
for (i = 1; i <= 9; i++) {
sum(muls[i - 1], b);
strcpy(muls[i], answer);
}

strcpy(quotient, "");
strcpy(remainder, "");
tryq = 1;

i = 0;
while (i != na) {
temp[0] = a[i];
strcat(remainder, temp);

tryq = 1;
//remainder>=muls[tryq]
while (tryq!=10 && !larger(muls[tryq], remainder))
tryq++;

temp[0] = tryq -1 + '0';
if (!(temp[0] == '0' && quotient[0] == '\0'))
strcat(quotient, temp);

sub(remainder, muls[tryq - 1]);
strcpy(remainder, answer);
if (remainder[0] == '0')
remainder[0] = '\0';

i++;
}
if (quotient[0] == '\0')
strcpy(quotient, "0");
if (remainder[0] == '\0')
strcpy(remainder, "0");

printf("%s\n", quotient);
}
int main(void)
{
char a[SIZE];
char b[SIZE];
char ch;
int na, nb;
int flaga, flagb;
while (1) {
na = 0; nb = 0;
flaga = 1; flagb = 1;
printf("A = ");
ch = getchar();
while (ch != '\n') {
if (ch != '-')
a[na++] = ch;
else
flaga = -1;
ch = getchar();
}
a[na] = '\0';

printf("B = ");
ch = getchar();
while (ch != '\n') {
if (ch != '-')
b[nb++] = ch;
else
flagb = -1;
ch = getchar();
}
b[nb] = '\0';

printf("A + B = ");
if (flaga == 1 && flagb == 1)
sum(a, b);
else if (flaga == 1 && flagb == -1)
sub(a, b);
else if (flaga == -1 && flagb == 1)
sub(b, a);
else if (flaga == -1 && flagb == -1) {
printf("-");
sum(a, b);
}
printf("%s\n", answer);

printf("A - B = ");
if (flaga == 1 && flagb == 1)
sub(a, b);
else if (flaga == 1 && flagb == -1)
sum(a, b);
else if (flaga == -1 && flagb == 1) {
printf("-");
sum(a, b);
}
else if (flaga == -1 && flagb == -1)
sub(b, a);
printf("%s\n", answer);

printf("A * B = ");
if (flaga == 1 && flagb == 1) {
if (a[0] == '0' || b[0] == '0')
printf("0\n");
else
mul(a, b);
}
else if (flaga == 1 && flagb == -1) {
if (a[0] == '0')
printf("0\n");
else {
printf("-");
mul(a, b);
}
}
else if (flaga == -1 && flagb == 1) {
if (b[0] == '0')
printf("0\n");
else {
printf("-");
mul(a, b);
}
}
else if (flaga == -1 && flagb == -1)
mul(a, b);

printf("A / B = ");
if (flaga == 1 && flagb == 1) {
if (b[0] == '0')
printf("Divide zero!!!\n");
else if (a[0] == '0')
printf("0\n");
else
divide(a, b);
}
else if (flaga == 1 && flagb == -1) {
if (a[0] == '0')
printf("0\n");
else {
printf("-");
divide(a, b);
}
}
else if (flaga == -1 && flagb == 1) {
if (b[0] == '0')
printf("Divide zero!!!\n");
else {
printf("-");
divide(a, b);
}
}
else if (flaga == -1 && flagb == -1)
divide(a, b);

printf("A %% B = ");
if (flagb == 1 && b[0] == '0')
printf("Divide zero!!!\n");
else if (flaga == 1 && a[0] == '0' || remainder[0] == '0')
printf("0\n");
else if (flaga == 1)
printf("%s\n", remainder);
else printf("-%s\n", remainder);

printf("=============\n");
}

getchar();
}

DH algorithm

How to use

All the operations happens in the folder “\SEC2014-HW2-3130000064-林涛\DH”

Choose prime

You should choose a prime and one of itscorrespondent primitive root, and put them in prime.txt and primitive_root.txtseparately.

I have choose a prime 300 numbers long asthe prime, which is recommended in “RFC2412”. You don’t have to do anythingabout it anymore.

Generate private key

Run “generate_private_key.exe”.



Figure 7 Generate private key
You can see a new file “private_key.txt” isbuild. The number in it is about 100 numbers long, and is generated randomly.



Figure8 private key

Generate public key

Run “generate_public_key.exe”.



Figure9 Generate public key
You can see a new file “public_key.txt” isbuild. The number in it is about 300 numbers long, and is generated accordingto prime.txt, private_key.txt and primitive_root.txt.



Figure10 public key

Generate secret key

Secret key is the key user actuallyconveyed, by user’s private key and the other’s public key.

You can put the private key and public keyinto a folder “A”, which means it’s the key for person A. And you can do thesame thing to a folder “B”.



Figure11 generate two pairs of keys
Put the private key of A and the public keyof B into the original folder and run “generate_secret_key.exe”.





Figure12 generate secret key
You can see a new file “public_secret.txt”is build. The number in it is about 300 numbers long, and is generated accordingto prime.txt, private_key.txt and public_key.txt.



Figure13 secret key (private A,public B)
Use the private key of B and the public keyof A to try again, you can see the same answer.



Figure14 secret key (private B,public A)

Algorithm

Just do as Diffie–Hellman algorithm,calculate A = ga mod
p.

It is impossible to calculate gadirectly when a is a large number, sowe should do “mod p” after eachmultiplication.

It is slow to multiply g for a times, we canreduce the time complexity intoO(log₂N)by diving into two parts each time.

Source code

generate_private_key.c

#include<stdio.h>
int main(void)
{
FILE *fp;
char prikey[200];
int i = 100;

srand((unsigned)time(NULL));

printf("Generate private key\n");

printf("Press ENTER");
getchar();

fp = fopen("private_key.txt","w");
fprintf(fp,"%c",rand()%9+'1');
while(i--)
fprintf(fp,"%c",rand()%10+'0');
if(fp != NULL)
printf("Generate private key successfully!\n");
fclose(fp);

}

generate_public_key.c

注：大量代码是与前面重复的，大概就只有main是新的。

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define SIZE 800
char answer[2 * SIZE];
char remainder[SIZE];
char quotient[SIZE];
//a,b都为自然数，且a>=b
void sum1(char *a, char *b)
{
int carry = 0;
int pa, pb;
int temp;
int i;
int na = strlen(a);
int nb = strlen(b);

answer[na] = '\0';
pa = na - 1; pb = nb - 1;
while (pa >= 0) {
temp = (a[pa] - '0') + carry;
if (pb >= 0)temp += (b[pb] - '0');

if (temp >= 10) {
answer[pa] = temp - 10 + '0';
carry = 1;
}
else {
answer[pa] = temp + '0';
carry = 0;
}
pa--; pb--;
}
if (carry) {
for (i = na; i >= 0; i--)
answer[i + 1] = answer[i];
answer[0] = '1';
}
}

void sub1(char *a, char *b)
{
int borrow = 0;
int pa, pb, pc;
int temp;
int i;
int na = strlen(a);
int nb = strlen(b);

answer[na] = '\0';
pa = na - 1; pb = nb - 1;
while (pa >= 0) {
temp = (a[pa] - '0') - borrow;
if (pb >= 0)temp -= (b[pb] - '0');

if (temp < 0) {
answer[pa] = temp + 10 + '0';
borrow = 1;
}
else {
answer[pa] = temp + '0';
borrow = 0;
}
pa--; pb--;
}
pc = 0;
while (answer[pc] == '0')
pc++;
for (i = 0; i <= na - pc; i++)
answer[i] = answer[i + pc];
}
//a,b都为自然数
void sum(char *a, char *b)
{
int na = strlen(a);
int nb = strlen(b);

if (na >= nb)
sum1(a, b);
else
sum1(b, a);
}

void sub(char *a, char *b)
{
int na = strlen(a);
int nb = strlen(b);

if (strcmp(a, b) == 0)
strcpy(answer, "0");
else if (na > nb || na == nb && strcmp(a, b) > 0)
sub1(a, b);
else {
printf("-");
sub1(b, a);
}
}

//a,b都是正整数
void mul(char *a, char *b)
{
char muls[10][SIZE + 1];
char temp[2 * SIZE];
int i, j;
int na = strlen(a);
int nb = strlen(b);

strcpy(muls[0], "0");

strcpy(temp, "0");
for (i = 1; i <= 9; i++) {
sum(muls[i - 1], a);
strcpy(muls[i], answer);
}

strcpy(answer, "0");

for (i = nb - 1; i >= 0; i--) {
if (b[i] != '0') {
strcpy(temp, muls[b[i]-'0']);
//后面加（nb-i-1）个零
for (j = 0; j < nb - i - 1; j++)
strcat(temp, "0");
sum(answer, temp);
}
}

//printf("%s\n", answer);

}
//a,b为自然数，若a>b返回1，不然0
int larger(char *a, char *b)
{
int na, nb;
na = strlen(a);
nb = strlen(b);
if (na > nb)
return 1;
else if (na == nb && strcmp(a, b) > 0)
return 1;
else
return 0;
}

void divide(char *a, char*b)
{
int na, nb; na = strlen(a); nb = strlen(b);
char temp[2]; temp[1] = '\0';
int tryq;
int i;

char muls[10][SIZE + 1];
strcpy(muls[0], "0");
strcpy(temp, "0");
for (i = 1; i <= 9; i++) {
sum(muls[i - 1], b);
strcpy(muls[i], answer);
}

strcpy(quotient, "");
strcpy(remainder, "");
tryq = 1;

i = 0;
while (i != na) {
temp[0] = a[i];
strcat(remainder, temp);

tryq = 1;
//remainder>=muls[tryq]
while (tryq!=10 && !larger(muls[tryq], remainder))
tryq++;

temp[0] = tryq -1 + '0';
if (!(temp[0] == '0' && quotient[0] == '\0'))
strcat(quotient, temp);

sub(remainder, muls[tryq - 1]);
strcpy(remainder, answer);
if (remainder[0] == '0')
remainder[0] = '\0';

i++;
}
if (quotient[0] == '\0')
strcpy(quotient, "0");
if (remainder[0] == '\0')
strcpy(remainder, "0");

}
/*
int main(void)
{
char a[SIZE];
char b[SIZE];
char ch;
int na, nb;
int flaga, flagb;
while (1) {
na = 0; nb = 0;
flaga = 1; flagb = 1;
printf("A = ");
ch = getchar();
while (ch != '\n') {
if (ch != '-')
a[na++] = ch;
else
flaga = -1;
ch = getchar();
}
a[na] = '\0';

printf("B = ");
ch = getchar();
while (ch != '\n') {
if (ch != '-')
b[nb++] = ch;
else
flagb = -1;
ch = getchar();
}
b[nb] = '\0';

printf("A + B = ");
if (flaga == 1 && flagb == 1)
sum(a, b);
else if (flaga == 1 && flagb == -1)
sub(a, b);
else if (flaga == -1 && flagb == 1)
sub(b, a);
else if (flaga == -1 && flagb == -1) {
printf("-");
sum(a, b);
}
printf("%s\n", answer);

printf("A - B = ");
if (flaga == 1 && flagb == 1)
sub(a, b);
else if (flaga == 1 && flagb == -1)
sum(a, b);
else if (flaga == -1 && flagb == 1) {
printf("-");
sum(a, b);
}
else if (flaga == -1 && flagb == -1)
sub(b, a);
printf("%s\n", answer);

printf("A * B = ");
if (flaga == 1 && flagb == 1) {
if (a[0] == '0' || b[0] == '0')
printf("0\n");
else
mul(a, b);
}
else if (flaga == 1 && flagb == -1) {
if (a[0] == '0')
printf("0\n");
else {
printf("-");
mul(a, b);
}
}
else if (flaga == -1 && flagb == 1) {
if (b[0] == '0')
printf("0\n");
else {
printf("-");
mul(a, b);
}
}
else if (flaga == -1 && flagb == -1)
mul(a, b);

printf("A / B = ");
if (flaga == 1 && flagb == 1) {
if (b[0] == '0')
printf("Devide zero!!!\n");
else if (a[0] == '0')
printf("0\n");
else
div(a, b);
}
else if (flaga == 1 && flagb == -1) {
if (a[0] == '0')
printf("0\n");
else {
printf("-");
div(a, b);
}
}
else if (flaga == -1 && flagb == 1) {
if (b[0] == '0')
printf("Devide zero!!!\n");
else {
printf("-");
div(a, b);
}
}
else if (flaga == -1 && flagb == -1)
div(a, b);

printf("A %% B = ");
if (flagb == 1 && b[0] == '0')
printf("Devide zero!!!\n");
else if (flaga == 1 && a[0] == '0')
printf("0\n");
else if (flaga == 1)
printf("%s\n", remainder);
else printf("-%s\n", remainder);

printf("=============\n");
}

getchar();
}*/

int main(void)
{
FILE *fp;
int i;
char prime[400];
char pubkey[800];
char prikey[200];
char binary[700];
char pr[10];

printf("Generate public key\n(According to prime.txt, private_key.txt and primitive_root.txt)\n");

fp = fopen("prime.txt","r");
if(fp == NULL){
printf("PLEASE INPUT prime.txt !!!\n");
return 0;
}
fscanf(fp,"%s",prime);
fclose(fp);

fp = fopen("private_key.txt","r");
if(fp == NULL){
printf("PLEASE INPUT private_key.txt !!!");
return 0;
}
fscanf(fp,"%s",prikey);
fclose(fp);

fp = fopen("primitive_root.txt","r");
if(fp == NULL){
printf("PLEASE INPUT primitive_root.txt !!!");
return 0;
}
fscanf(fp,"%s",pr);
fclose(fp);

printf("Press ENTER");
getchar();

i=0;
while(prikey[0]!='0'){
divide(prikey,"2");
binary[i++] = remainder[0];
strcpy(prikey, quotient);
}
binary[i] = 0;

strcpy(pubkey, "1");

for(i--;i>=0;i--){
mul(pubkey,pubkey);
strcpy(pubkey, answer);
if(binary[i] == '1'){
mul(pubkey,pr);
strcpy(pubkey, answer);
}
divide(pubkey, prime);
strcpy(pubkey, remainder);
printf("%s\n",pubkey);
}

fp = fopen("public_key.txt","w");
fprintf(fp,"%s",pubkey);
if(fp != NULL)
printf("Generate public key successfully!\n");
fclose(fp);

}

generate_secret_key.c

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define SIZE 800
char answer[2 * SIZE];
char remainder[SIZE];
char quotient[SIZE];
//a,b都为自然数，且a>=b
void sum1(char *a, char *b)
{
int carry = 0;
int pa, pb;
int temp;
int i;
int na = strlen(a);
int nb = strlen(b);

answer[na] = '\0';
pa = na - 1; pb = nb - 1;
while (pa >= 0) {
temp = (a[pa] - '0') + carry;
if (pb >= 0)temp += (b[pb] - '0');

if (temp >= 10) {
answer[pa] = temp - 10 + '0';
carry = 1;
}
else {
answer[pa] = temp + '0';
carry = 0;
}
pa--; pb--;
}
if (carry) {
for (i = na; i >= 0; i--)
answer[i + 1] = answer[i];
answer[0] = '1';
}
}

void sub1(char *a, char *b)
{
int borrow = 0;
int pa, pb, pc;
int temp;
int i;
int na = strlen(a);
int nb = strlen(b);

answer[na] = '\0';
pa = na - 1; pb = nb - 1;
while (pa >= 0) {
temp = (a[pa] - '0') - borrow;
if (pb >= 0)temp -= (b[pb] - '0');

if (temp < 0) {
answer[pa] = temp + 10 + '0';
borrow = 1;
}
else {
answer[pa] = temp + '0';
borrow = 0;
}
pa--; pb--;
}
pc = 0;
while (answer[pc] == '0')
pc++;
for (i = 0; i <= na - pc; i++)
answer[i] = answer[i + pc];
}
//a,b都为自然数
void sum(char *a, char *b)
{
int na = strlen(a);
int nb = strlen(b);

if (na >= nb)
sum1(a, b);
else
sum1(b, a);
}

void sub(char *a, char *b)
{
int na = strlen(a);
int nb = strlen(b);

if (strcmp(a, b) == 0)
strcpy(answer, "0");
else if (na > nb || na == nb && strcmp(a, b) > 0)
sub1(a, b);
else {
printf("-");
sub1(b, a);
}
}

//a,b都是正整数
void mul(char *a, char *b)
{
char muls[10][SIZE + 1];
char temp[2 * SIZE];
int i, j;
int na = strlen(a);
int nb = strlen(b);

strcpy(muls[0], "0");

strcpy(temp, "0");
for (i = 1; i <= 9; i++) {
sum(muls[i - 1], a);
strcpy(muls[i], answer);
}

strcpy(answer, "0");

for (i = nb - 1; i >= 0; i--) {
if (b[i] != '0') {
strcpy(temp, muls[b[i]-'0']);
//后面加（nb-i-1）个零
for (j = 0; j < nb - i - 1; j++)
strcat(temp, "0");
sum(answer, temp);
}
}

//printf("%s\n", answer);

}
//a,b为自然数，若a>b返回1，不然0
int larger(char *a, char *b)
{
int na, nb;
na = strlen(a);
nb = strlen(b);
if (na > nb)
return 1;
else if (na == nb && strcmp(a, b) > 0)
return 1;
else
return 0;
}

void divide(char *a, char*b)
{
int na, nb; na = strlen(a); nb = strlen(b);
char temp[2]; temp[1] = '\0';
int tryq;
int i;

char muls[10][SIZE + 1];
strcpy(muls[0], "0");
strcpy(temp, "0");
for (i = 1; i <= 9; i++) {
sum(muls[i - 1], b);
strcpy(muls[i], answer);
}

strcpy(quotient, "");
strcpy(remainder, "");
tryq = 1;

i = 0;
while (i != na) {
temp[0] = a[i];
strcat(remainder, temp);

tryq = 1;
//remainder>=muls[tryq]
while (tryq!=10 && !larger(muls[tryq], remainder))
tryq++;

temp[0] = tryq -1 + '0';
if (!(temp[0] == '0' && quotient[0] == '\0'))
strcat(quotient, temp);

sub(remainder, muls[tryq - 1]);
strcpy(remainder, answer);
if (remainder[0] == '0')
remainder[0] = '\0';

i++;
}
if (quotient[0] == '\0')
strcpy(quotient, "0");
if (remainder[0] == '\0')
strcpy(remainder, "0");

}
/*
int main(void)
{
char a[SIZE];
char b[SIZE];
char ch;
int na, nb;
int flaga, flagb;
while (1) {
na = 0; nb = 0;
flaga = 1; flagb = 1;
printf("A = ");
ch = getchar();
while (ch != '\n') {
if (ch != '-')
a[na++] = ch;
else
flaga = -1;
ch = getchar();
}
a[na] = '\0';

printf("B = ");
ch = getchar();
while (ch != '\n') {
if (ch != '-')
b[nb++] = ch;
else
flagb = -1;
ch = getchar();
}
b[nb] = '\0';

printf("A + B = ");
if (flaga == 1 && flagb == 1)
sum(a, b);
else if (flaga == 1 && flagb == -1)
sub(a, b);
else if (flaga == -1 && flagb == 1)
sub(b, a);
else if (flaga == -1 && flagb == -1) {
printf("-");
sum(a, b);
}
printf("%s\n", answer);

printf("A - B = ");
if (flaga == 1 && flagb == 1)
sub(a, b);
else if (flaga == 1 && flagb == -1)
sum(a, b);
else if (flaga == -1 && flagb == 1) {
printf("-");
sum(a, b);
}
else if (flaga == -1 && flagb == -1)
sub(b, a);
printf("%s\n", answer);

printf("A * B = ");
if (flaga == 1 && flagb == 1) {
if (a[0] == '0' || b[0] == '0')
printf("0\n");
else
mul(a, b);
}
else if (flaga == 1 && flagb == -1) {
if (a[0] == '0')
printf("0\n");
else {
printf("-");
mul(a, b);
}
}
else if (flaga == -1 && flagb == 1) {
if (b[0] == '0')
printf("0\n");
else {
printf("-");
mul(a, b);
}
}
else if (flaga == -1 && flagb == -1)
mul(a, b);

printf("A / B = ");
if (flaga == 1 && flagb == 1) {
if (b[0] == '0')
printf("Devide zero!!!\n");
else if (a[0] == '0')
printf("0\n");
else
div(a, b);
}
else if (flaga == 1 && flagb == -1) {
if (a[0] == '0')
printf("0\n");
else {
printf("-");
div(a, b);
}
}
else if (flaga == -1 && flagb == 1) {
if (b[0] == '0')
printf("Devide zero!!!\n");
else {
printf("-");
div(a, b);
}
}
else if (flaga == -1 && flagb == -1)
div(a, b);

printf("A %% B = ");
if (flagb == 1 && b[0] == '0')
printf("Devide zero!!!\n");
else if (flaga == 1 && a[0] == '0')
printf("0\n");
else if (flaga == 1)
printf("%s\n", remainder);
else printf("-%s\n", remainder);

printf("=============\n");
}

getchar();
}*/

int main(void)
{
FILE *fp;
int i;
char prime[400];
char pubkey[400];
char prikey[200];
char binary[700];
char secretkey[1200];

printf("Gnerate secret key\n(According to prime.txt, private_key.txt and public_key.txt)\n");

fp = fopen("prime.txt","r");
if(fp == NULL){
printf("PLEASE INPUT prime.txt !!!\n");
return 0;
}
fscanf(fp,"%s",prime);
fclose(fp);

fp = fopen("private_key.txt","r");
if(fp == NULL){
printf("PLEASE INPUT private_key.txt !!!");
return 0;
}
fscanf(fp,"%s",prikey);
fclose(fp);

fp = fopen("public_key.txt","r");
if(fp == NULL){
printf("PLEASE INPUT public_key.txt !!!");
return 0;
}
fscanf(fp,"%s",pubkey);
fclose(fp);

printf("Press ENTER");
getchar();

i=0;
while(prikey[0]!='0'){
divide(prikey,"2");
binary[i++] = remainder[0];
strcpy(prikey, quotient);
}
binary[i] = 0;

strcpy(secretkey, "1");

for(i--;i>=0;i--){
mul(secretkey,secretkey);
strcpy(secretkey, answer);
if(binary[i] == '1'){
mul(secretkey,pubkey);
strcpy(secretkey, answer);
}
divide(secretkey, prime);
strcpy(secretkey, remainder);
printf("%s\n",secretkey);
}

fp = fopen("secret_key.txt","w");
fprintf(fp,"%s",secretkey);
if(fp != NULL)
printf("Generate secret key successfully!\n");
fclose(fp);

}

prime.txt

179769313486231590770839156793787453197860296048756011706444423684197180216158519368947833795864925541502180565485980503646440548199239100050792877003355816639229553136239076508735759914822574862575007425302077447712589550957937778424442426617334727629299387668709205606050270810842907692932019128194467627007

primitive_root.txt

2