超大整数运算算法——为RSA加密算法提供运算工具
/* program: Large integer operations
* Made by: Daiyyr
* date: 2013/07/09
* This software is licensed under the terms of the GNU General Public
* License version 2, as published by the Free Software Foundation, and
* may be copied, distributed, and modified under those terms.
*/
//开源代码,引用请务必遵守GNU规则
//未完待续,尚缺 乘方、除法、取模运算
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
//#define DEBUG
#ifdef DEBUG
#define LOG(format, args...) do { printf(format, ## args); \
printf("\n");}\
while(0)
#define DBG() printk("[%s]:%d => \n",__FUNCTION__,__LINE__)
#else
#define LOG(format, args...);
#define DBG()
#endif
#define LOGE(format, args...) do { printf(format, ## args); \
printf("\n");}\
while(0)
#define LENGTH 1000 //max digit of this long number
struct charnumber{
char sign; //0 for positive, 1 for negative
int length;
unsigned char* array;//low index of this array is for high digit of the number
unsigned char* pointerforfree;
};
int char2interger(unsigned char onebyte){
if (onebyte < 10)
return onebyte;
else
return ((onebyte>>4)*10)+(onebyte&15);
}
struct charnumber* interger2charnumber(signed long int number){
unsigned char charnumber[LENGTH/2];
int i=0;
struct charnumber *target;
LOG("num: %ld", number);
target = malloc(sizeof(struct charnumber));
if (number<0){
target->sign=1;
number = -number;
LOG("num:%ld", number);
}
else{
target->sign=0;
}
while(number!=0){
charnumber[i]=number%10;
LOG("i:%d, num:%ld, char:%d", i, number, charnumber[i]);
number/=10;
charnumber[i]|=number%10<<4;
LOG("i:%d, num:%ld, char:%d", i, number, charnumber[i]);
number/=10;
i++;
}
target->length=i;
target->array = malloc(target->length);
target->pointerforfree = target->array;
for(i--;i>(target->length-1)/2;i--){ //Reverse
LOG("length=%d, i=%d, left=%d, right=%d", target->length, i, charnumber[target->length-i-1], charnumber[i]);
charnumber[i] = charnumber[i] + charnumber[target->length-i-1];
charnumber[target->length-i-1] = charnumber[i] - charnumber[target->length-i-1];
charnumber[i] = charnumber[i] - charnumber[target->length-i-1];
}
memcpy(target->array, charnumber, target->length);
for(i=0;i<target->length;i++)
LOG("i:%d:%d", i, char2interger(target->array[i]));
LOG("~~length:%d", target->length, target->length);
return target;
}
char* charnumber2string(struct charnumber* charnumber){
int length = 2*charnumber->length+1;
char* showchar = malloc(length);
int i;
LOG("string signed:%d", charnumber->sign);
if(charnumber->sign!=0)
showchar[0]='-';
else
showchar[0]='+';
for(i=0; i<charnumber->length; i++){
showchar[2*i+1]='0'+(charnumber->array[i]>>4);
showchar[2*i+2]='0'+(charnumber->array[i]&15);
LOG("%d: %x, %c%c",i, charnumber->array[i], showchar[2*i+1],showchar[2*i+2]);
}
showchar[length]='\0';
if (!strcmp(showchar, "+"))
strcpy(showchar, "+0");
else if(showchar[1]=='0')
showchar++, showchar[0]=charnumber->sign ? '-' : '+';
return showchar;
}
struct charnumber* string2charnumber(char* string){
if(!(string[0]=='-' || string[0]=='+' || (string[0]>='0' && string[0]<='9'))){
printf("error: in string2charnumber, invalid input number!\n");
return NULL;
}
struct charnumber *number;
char* temp = malloc(strlen(string));
int i, odd;
strcpy(temp, string);
number = malloc(sizeof(struct charnumber));
if(temp[0]=='-'){
number->sign = 1;
temp++;
}
else{
number->sign = 0;
}
odd = strlen(temp)%2;
LOG("strlen(temp): %d, odd=%d\n", strlen(temp), odd);
number->length = strlen(temp)/2 + odd;
number->array = malloc(number->length);
number->pointerforfree = number->array;
for(i=number->length-1; i>=0; i--){
if(i){
if(temp[i*2+1-odd] > '9' || temp[i*2+1-odd] < '0' || temp[i*2-odd] > '9' || temp[i*2-odd] < '0'){
printf("error! in string2charnumber, invalid input number!\n");
LOG("temp[%d]=%x, temp[%d]=%x",i*2+1-odd, temp[i*2+1-odd], i*2-odd, temp[i*2-odd]);
return NULL;
}
number->array[i]=(temp[i*2+1-odd] - '0') + ((temp[i*2-odd] - '0') << 4);
LOG("number->array[%d]=%x", i, number->array[i]);
}
else{
number->array[0]=odd ? temp[0] - '0' : (temp[1] - '0') + ((temp[0] - '0') << 4);
LOG("temp[1]:%d, temp[0]:%d, '0':%d, number->array[%d]=%x",temp[1],temp[0], '0', i, number->array[i]);
}
}
return number;
}
struct charnumber* minusCharNumber(struct charnumber* charnumber1, struct charnumber* charnumber2);
struct charnumber* plusCharNumber(struct charnumber* charnumber1, struct charnumber* charnumber2){
struct charnumber *temp1, *temp2, *sum;
if(!!charnumber1->sign != !!charnumber2->sign){
temp1 = malloc(sizeof(struct charnumber));
temp2 = malloc(sizeof(struct charnumber));
memcpy(temp1, charnumber1, sizeof(struct charnumber));
memcpy(temp2, charnumber2, sizeof(struct charnumber));
temp2->sign = !temp2->sign;
sum = minusCharNumber(temp1, temp2);
free(temp1->pointerforfree), free(temp2->pointerforfree), free(temp1), free(temp2);
return sum;
}
int i, carry = 0;
if(charnumber1->length >= charnumber2->length){
temp1 = charnumber1;
temp2 = charnumber2;
}
else{
temp1 = charnumber2;
temp2 = charnumber1;
}
sum = malloc(sizeof(struct charnumber));
sum->sign = temp1->sign;
LOG("signed:%d", sum->sign);
sum->array = malloc(temp1->length+1);
sum->pointerforfree = sum->array;
for(i=temp1->length; i>temp1->length-temp2->length; i--){ //遍历较小数的位数
sum->array[i]=(carry + temp1->array[i-1]&15) + (temp2->array[i-(temp1->length+1-temp2->length)]&15);
carry = 0;
if(sum->array[i]>9){
sum->array[i] = 6 + sum->array[i];
}
sum->array[i]=sum->array[i] + (temp1->array[i-1]>>4<<4) + (temp2->array[i-(temp1->length+1-temp2->length)]>>4<<4);
if(sum->array[i]>159)
sum->array[i] -= 160, carry = 1;
LOG("i: %d, 0x%x, 1:0x%x, 2:0x%x", i, sum->array[i], temp1->array[i-1], temp2->array[i-(temp1->length+1-temp2->length)]);
}
while(carry && i>=1){ //在遍历完较小数的位数后,处理可能的进位
sum->array[i] = temp1->array[i-1] + 1;
if((sum->array[i]&15) == 10){
LOG("carry1.0, %x", sum->array[i]);
sum->array[i]=(sum->array[i]&240) + 16;
LOG("carry1.1, %x", sum->array[i]);
if(sum->array[i] == 160){
sum->array[i]=0;
carry=1;
LOG("carry2");
}
else
carry=0;
}
else
carry=0;
i--;
}
if(carry){
sum->array[0]=1;
sum->length = temp1->length+1;
}
else{
sum->array++;
memcpy(sum->array, temp1->array, i);
sum->length = temp1->length;
}
return sum;
}
struct charnumber* multiplyCharNumberAndInt(struct charnumber* charnumber, int n){
long i;
struct charnumber *result, *temp;
result = interger2charnumber(0);
for(i=0; i<n; i++){
temp = result;
result = plusCharNumber(result, charnumber);
free(temp->pointerforfree);
free(temp);
}
if(charnumber->sign ^ (n<0) )
result->sign = 1;
else
result->sign = 0;
return result;
}
int mypow(int x, int y){
int i, sum=1;
for(i=0; i<y; i++){
sum=sum*x;
}
return sum;
}
struct charnumber* multiplyCharNumber(struct charnumber* charnumber1, struct charnumber* charnumber2){
int i, orisign1, orisign2;
struct charnumber* result, *temp1, *temp2, *tempforfree1, *tempforfree2, *tempforfree3;
orisign1 = charnumber1->sign;
orisign2 = charnumber2->sign;
charnumber1->sign = 0;
charnumber2->sign = 0;
if(charnumber1->length >= charnumber2->length){
temp1 = charnumber1;
temp2 = charnumber2;
}
else{
temp1 = charnumber2;
temp2 = charnumber1;
}
result = interger2charnumber(0);
LOG("t1length:%d, t2length:%d", temp1->length, temp2->length);
for(i=0; i<temp2->length; i++){
tempforfree1 = multiplyCharNumberAndInt(temp1, temp2->array[i]&15);
tempforfree2 = multiplyCharNumberAndInt(tempforfree1, mypow(10, (temp2->length-i)*2-2));
tempforfree3 = result;
result = plusCharNumber(tempforfree3, tempforfree2);
free(tempforfree1->pointerforfree), free(tempforfree2->pointerforfree), free(tempforfree3->pointerforfree), free(tempforfree1), free(tempforfree2), free(tempforfree3);
LOG("multip1 i=%d, t1=0x%x, t2=0x%x, t3=0x%x, res=0x%x", i, tempforfree1->array[i], tempforfree2->array[i], tempforfree3->array[i], result->array[i]);
tempforfree1 = multiplyCharNumberAndInt(temp1, temp2->array[i]>>4);
tempforfree2 = multiplyCharNumberAndInt(tempforfree1, mypow(10, (temp2->length-i)*2-1));
LOG("tempforfree1:%s, tempforfree2:%s",charnumber2string(tempforfree1), charnumber2string(tempforfree2));
tempforfree3 = result;
result = plusCharNumber(tempforfree3, tempforfree2);
free(tempforfree1->pointerforfree), free(tempforfree2->pointerforfree), free(tempforfree3->pointerforfree), free(tempforfree1), free(tempforfree2), free(tempforfree3);
}
charnumber1->sign = orisign1;
charnumber2->sign = orisign2;
if(charnumber1->sign ^ charnumber2->sign ){
result->sign = 1;
}
else
result->sign = 0;
return result;
}
struct charnumber* minusCharNumber(struct charnumber* charnumber1, struct charnumber* charnumber2){
struct charnumber *temp1, *temp2, *difference;
if(!!charnumber1->sign != !!charnumber2->sign){
temp1 = malloc(sizeof(struct charnumber));
temp2 = malloc(sizeof(struct charnumber));
memcpy(temp1, charnumber1, sizeof(struct charnumber));
memcpy(temp2, charnumber2, sizeof(struct charnumber));
temp2->sign = !temp2->sign;
difference = plusCharNumber(temp1, temp2);
free(temp1->pointerforfree), free(temp2->pointerforfree), free(temp1), free(temp2);
return difference;
}
int i, borrow = 0;
difference = malloc(sizeof(struct charnumber));
if(charnumber1->length > charnumber2->length){
temp1 = charnumber1;
temp2 = charnumber2;
difference->sign = temp1->sign;
}
else if(charnumber1->length < charnumber2->length){
temp1 = charnumber2;
temp2 = charnumber1;
difference->sign = -temp1->sign;
}
else{
for(i=0; i<charnumber1->length; i++){
if(charnumber1->array[i] > charnumber2->array[i]){
temp1 = charnumber1;
temp2 = charnumber2;
difference->sign = temp1->sign;
break;
}
else if(charnumber1->array[i] < charnumber2->array[i]){
temp1 = charnumber2;
temp2 = charnumber1;
difference->sign = -temp1->sign;
break;
}
}
if(i == charnumber1->length)
return interger2charnumber(0);
}
difference->array = malloc(temp1->length);
difference->pointerforfree = difference->array;
for(i=temp1->length-1; i>=temp1->length-temp2->length; i--){ //遍历较小数的位数
difference->array[i]=(borrow + temp1->array[i]&15) - (temp2->array[i-(temp1->length-temp2->length)]&15);
if(difference->array[i]>10){
difference->array[i] = difference->array[i]-6;
}
LOG("minus:i: %d, 0x%x, 1:0x%x, 2:0x%x", i, difference->array[i], temp1->array[i], temp2->array[i-(temp1->length-temp2->length)]);
borrow = 0;
difference->array[i]=difference->array[i] + (temp1->array[i]>>4<<4) - (temp2->array[i-(temp1->length-temp2->length)]>>4<<4);
if(difference->array[i]>160)
difference->array[i] += 160, borrow = -1;
LOG("minus:i: %d, 0x%x, 1:0x%x, 2:0x%x", i, difference->array[i], temp1->array[i], temp2->array[i-(temp1->length-temp2->length)]);
}
if(!borrow && temp1->length - temp2->length == 1)
memcpy(difference->array, temp1->array, i+1);
while(borrow && i>=0){ //在遍历完较小数的位数后,处理可能的借位
difference->array[i] = temp1->array[i] - 1;
if((difference->array[i]&15) == 15){
LOG("carry1.0, %x", difference->array[i]);
difference->array[i]=(difference->array[i]&240) - 7;
LOG("carry1.1, %x", difference->array[i]);
if(difference->array[i] == 249){//0xf9
difference->array[i]=153;//0x99
borrow=-1;
LOG("carry2");
}
else
borrow=0;
}
else
borrow=0;
i--;
}
LOG("BEFORE WHILE, i=%d", i);
if(i>0){
LOG("memcpy");
memcpy(difference->array, temp1->array, i+1);
difference->length = temp1->length;
}
else{
i=0;
while(difference->array[0]==0){
difference->array++;
i++;
}
difference->length = temp1->length-i;
}
return difference;
}
//struct charnumber* powerCharNumber(struct charnumber* charnumber1, struct charnumber* pow){
// struct charnumber* char1, sum=1;
// int i;
// char1 = malloc(struct charnumber);
// char1->length = charnumber1->length;
// char1->sign = charnumber1->sign;
// char1->array = malloc(char1->length);
// char1->pointerforfree = char1->array;
// memcpy(char1->array, charnumber1->array, char1->length);
// while(char1->length != 0){
// if(char1->array[char1->length-1]&15 == 0){
// if(char1->array[char1->length-1] == 0){
// char1->array[char1->length-2]
// }
// }
// sum=multiplyCharNumber(sum, charnumber1);
// char1->length--;
// }
// return sum;
//}
超大整数运算算法——为RSA加密算法提供运算工具的更多相关文章
- 数据加密--详解 RSA加密算法 原理与实现
RSA算法简介 RSA是最流行的非对称加密算法之一.也被称为公钥加密.它是由罗纳德·李维斯特(Ron Rivest).阿迪·萨莫尔(Adi Shamir)和伦纳德·阿德曼(Leonard Adlema ...
- RSA加密算法的加密与解密
转发原文链接:RSA加密算法加密与解密过程解析 1.加密算法概述 加密算法根据内容是否可以还原分为可逆加密和非可逆加密. 可逆加密根据其加密解密是否使用的同一个密钥而可以分为对称加密和非对称加密. 所 ...
- RSA加密算法 C++实现
上信息安全课,老师布置了几个大作业,其中一个为RSA加密算法的实现,不能用Java写.出于兴趣,决定尝试.完成之后,为了便于查找,于是写下这篇文章,以备后续查看.也供大家一起学习,一起进步. 1.预备 ...
- 用实例讲解RSA加密算法(精)
RSA是第一个比较完善的公开密钥算法,它既能用于加密,也能用于数字签名.RSA以它的三个发明者Ron Rivest, Adi Shamir, Leonard Adleman的名字首字母命名,这个算法经 ...
- 轻松学习RSA加密算法原理
转自:http://blog.csdn.net/sunmenggmail/article/details/11994013 http://blog.csdn.net/q376420785/articl ...
- RSA加密算法原理及RES签名算法简介
第一部分:RSA算法原理与加密解密 一.RSA加密过程简述 A和B进行加密通信时,B首先要生成一对密钥.一个是公钥,给A,B自己持有私钥.A使用B的公钥加密要加密发送的内容,然后B在通过自己的私钥解密 ...
- 轻松学习RSA加密算法原理 (转)
轻松学习RSA加密算法原理 (转) http://blog.csdn.net/q376420785/article/details/8557266 http://www.ruanyifeng.com/ ...
- [转] 用实例给新手讲解RSA加密算法
http://www.cfca.com.cn/zhishi/wz-012.htm PS: 通常公钥对数据加密,私钥对数据解密:私钥对数据签名,公钥对数据签名进行认证 RSA加密算法是最常用的非对称加密 ...
- 用实例给新手讲解RSA加密算法
图为 RSA公开密钥算法的发明人,从左到右Ron Rivest, Adi Shamir, Leonard Adleman. 照片摄于1978年 RSA加密算法是最常用的非对称加密算法,CFCA在证书服 ...
随机推荐
- 数据导出为excel表格
---恢复内容开始--- 方式一: 通过request和response中携带的数据导出表格,导出的结果会将页面中展示的内容全部导出.代码如下: //调出保存框,下载页面所有内容 String fil ...
- DepthClipEnable error
刚刚呢又遇到这种鬼扯的问题,ps就return个(1,1,0,1) nisight的汇编都写对了,但结果就是画不出任何东西,按照经验,必然是某个state.... 我就找啊找啊,被我找到一个 Rast ...
- webservice之XFire的使用(java调用java)
注意:xfire不支持java.util.List等集合,所以调用webservice传递的参数要为基本类型. 转自:http://zp9245.blog.163.com/blog/static/10 ...
- javascript document.write
在载人页面后,浏览器输出流自动关闭:在此之后,任何一个对当前页面进行操作的document.write()方法将打开—个新的输出流.它将清除当前页面内容(包括源文档的任何变量或值).document. ...
- Grid分组汇总
Ext.onReady(function () { Ext.define('personInfo', { extend: 'Ext. ...
- ExtJs之字段集FieldSet
//Ext.form.FieldSet扩展自Ext.container.Container.其优点就是把相同字段集中在一起,在外面字段外面加了个线"围住"他们. // ...
- vi/vim使用指北 ---- Introducting the ex Editor
本章介绍ex编辑器,为什么要介绍这样一个新的编辑器呢:其实ex编辑器不能算是一个新的编辑器,vi只是它的visual model,它已经是一个更普遍,基于行的编辑器.ex提供更大机动和范围的编辑命令. ...
- ios下最简单的正则,RegexKitLite
ios下最简单的正则,RegexKitLite 1.去RegexKitLite下载类库,解压出来会有一个例子包及2个文件,其实用到的就这2个文件,添加到工程中.备用地址:http://www.coco ...
- Flatty Shadow在线为Icon图标生成长阴影效果。
Flatty Shadow在线为Icon图标生成长阴影效果. Flatty Shadow 彩蛋爆料直击现场 Flatty Shadow在线为Icon图标生成长阴影效果.
- 使用动态SQL语句实现简单的行列转置(动态产生列)
原始数据如下图所示:(商品的销售明细)date=业务日期:Item=商品名称:saleqty=销售数量: -- 建立测试数据(表)create table test (Date varchar(10) ...