密码学课程设计(格式)2016

1、南京航空航天大学课 程 设 计 报 告课 程 名 称 密码学课程设计 学 院 计算机科学与技术 年 级 2014 学 生 姓 名 陶超权 学 号 161420330 开 课 时 间 2016 至 2017 学年第 一 学期总 成 绩教师签名实验项目名 称实验一、古典密码成绩一、实验目的通过实现简单的古典密码算法，理解密码学的相关概念如明文（plaintext）、密文（ciphertext）、加密密钥（encryption key）、解密密钥（decryption key）、加密算法(encryption algorithm)、解密算法（decryption algorithm）等。2、 实验内

2、容1）用CC+语言实现单表仿射（Affine）加/解密算法；2）用CC+语言实现统计26个英文字母出现频率的程序；3）利用单表仿射加/解密程序对一段较长的英文文章进行加密，再对明文和密文中字母出现的频率进行统计并作对比，观察有什么规律。 仿射变换：加密：解密：其中，k1和k2为密钥，k1Zq，k2Zq*。3、 实验步骤1）在main函数中构建框架，函数主要分为三部分，加密，解密，计算字符出现频率；2）加密函数encrypt(),首先需要输入两个密钥K1，k2,需要注意k2是和26互质的，所以这里用gcd()函数判断了k2与26的最大公约数，加解密都采用了文件操作，明文和密文都保存在文件中，这里

3、加密时根据ascii码，对大小字母分别加密，其他字符则保持不变；3）解密函数decode(),和加密函数类似，需要注意解密要用到密钥K2的逆元，所以这里用函数inverse_k2()进行了逆元的求解，另外需要注意的是解密运算过程中可能出现数值为负数的情况，在模运算下应该将它们重新置为整数。4）计算字符频率函数calculateCharFreq()，这里只对大小字母进行统计，不计其他字符。源代码：* main.cpp *#include<stdio.h>#include<stdlib.h>int main() void encrypt(); void decode();

4、void calculateCharFreq(); int choice; printf("please input your choice:n"); printf("t1. encryptnt2.decodent3.calculate character frequencent4.quit&exitn"); scanf("%d",&choice); getchar(); while(1) switch(choice) case 1: encrypt(); break; case 2: decode(); break;

5、 case 3: calculateCharFreq(); break; case 4: return 0; default:break; printf("please input your choice:n"); printf("t1. encryptnt2.decodent3.calculate character frequencent4.quit&exitn"); scanf("%d",&choice); getchar(); return 0;* encrypt.cpp *#include<stdlib

6、.h>#include<stdio.h>void encrypt() int gcd(int,int); printf("please input secret keyn"); printf("k1:"); int k1,k2; scanf("%d",&k1); getchar(); while(k1<0 | k1>=26) printf("illegal k1!please input agian!n"); printf("k1:"); scanf(&qu

7、ot;%d",&k1); getchar(); printf("k2:"); scanf("%d",&k2); getchar(); while(gcd(k2,26)!=1) printf("illegal k2!please input agian!"); printf("k2:"); scanf("%d",&k2); getchar(); / char plainText100='0' / char cypherText100='0&

8、#39; printf("open the plain text filen"); /scanf("%s",plainText); FILE *fplain,*fcypher; if(fplain =fopen("plain.txt","a+")=NULL) printf("can't open plain.txt!n'"); exit(0); if(fcypher = fopen("cypher.txt","a")=NULL) print

9、f("can't open cypher.txt!n'"); exit(0); /int i; char plainText = fgetc(fplain); char cypherText; /for(i=0; plainTexti!='0' i+) / printf("%c",plainText); while(plainText != EOF) / printf("old:%dn",plainTexti); if(65<=plainText&& plainText<=9

10、0 ) / printf("."); / plainTexti-=65; cypherText = (plainText-65)*k2+k1)%26+65; fputc(cypherText,fcypher); / printf("new:%dn",cypherTexti); else if(97<=plainText && plainText<=122 ) /plainTexti-=97; cypherText = (plainText-97)*k2+k1)%26+97; fputc(cypherText,fcypher);

11、 / printf("new:%dn",cypherTexti); else cypherText=plainText; fputc(cypherText,fcypher); plainText = fgetc(fplain); / printf("%ct%c",plainTexti,cypherTexti); fclose(fplain); fclose(fcypher); printf("cipher text has been written into cypher.txt!n"); / printf("%snn&qu

12、ot;,cypherText);* decode.cpp *#include<stdlib.h>#include<stdio.h>void decode() int gcd(int ,int); int inverse(int);/ char cypherText100='0'/ char plainText200='0' printf("please input secret keyn"); printf("k1:"); int k1,k2; scanf("%d",&k

13、1); getchar(); while(k1<0 | k1>=26) printf("illegal k1!please input agian!n"); printf("k1:"); scanf("%d",&k1); getchar(); printf("k2:"); scanf("%d",&k2); getchar(); while(gcd(k2,26)!=1) printf("illegal k2!please input agian!")

14、; printf("k2:"); scanf("%d",&k2); getchar(); int inverse_k2 = inverse(k2); printf("open the cypher txt!n");/ scanf("%s",cypherText); FILE *fplain,*fcypher; if(fplain =fopen("plain1.txt","a")=NULL) printf("can't open plain.txt!n

15、'"); exit(0); if(fcypher = fopen("cypher.txt","r")=NULL) printf("can't open cypher.txt!n'"); exit(0); /int i; char cypherText = fgetc(fcypher); char plainText; /for(i=0; plainTexti!='0' i+) / printf("%c",plainText);/ int i; / for(i=0;

16、cypherTexti!='0' i+) while(cypherText!=EOF) / printf("old:%dn",plainTexti); if(65<=cypherText&& cypherText<=90 ) / printf("."); / plainTexti-=65; int t= cypherText-65-k1 ; if(t<0) int i=1; while(t<0) t+=26 * i ; i+; plainText = (t*inverse_k2)%26)+65; f

17、putc(plainText,fplain); / printf("new:%dn",cypherTexti); else if(97<=cypherText && cypherText<=122 ) /plainTexti-=97; int t= cypherText-97-k1 ; if(t<0) int i=1; while(t<0) t+=26 * i ; i+; plainText = (t*inverse_k2)%26)+97; / printf("new:%dn",plainText); fputc(

18、plainText,fplain); else plainText=cypherText; fputc(plainText,fplain); /printf("%ct%c",plainTexti,cypherTexti); cypherText= fgetc(fcypher); fclose(fplain); fclose(fcypher); printf("plain text has been written into plain1.txt!n"); / printf("%snn",plainText);int gcd(int x

19、, int y) int t=0; /int t1=x; /int t2=y; while(y!=0) t = x%y; x = y; y = t; /printf("%d",x); return x;int inverse(int x) int i; for( i=0; i<26; i+) if(i*x)%26=1) break; / printf("%dn",i); return i;* charFreq.cpp *#include<stdio.h>#include<stdlib.h>void calculateChar

20、Freq() char filename10='0' char ch; FILE * fp; float sum=0; float freq26=0; float charNum26=0; printf("please input the filename you wanna staticstic frequence:n "); /getchar(); gets(filename); /getchar(); if(fp =fopen(filename,"r")=NULL) printf("can't open %s!n&

21、quot;,filename); exit(0); ch = fgetc(fp); while(ch != EOF) if('a'<=ch &&ch<='z') charNumch-97+; sum+; /printf("charNum:%ftsum:%fn",charNumch-97,sum); else if(ch>='A' && ch<='Z') charNumch-65+; sum+; /printf("%cn",ch); ch

22、=fgetc(fp); fclose(fp); /char t = 0; int count=0; for(int i=0;i<26;i+) freqi = charNumi/sum; /printf("charNUm:%ftsum:%ftfreq:%fn",charNumi,sum,freqi); printf("%c(%c):%.3ft",i+65,i+97,freqi); count+; if(count=6) printf("n"); count=0; printf("nn");四、实验结果及分析这里

23、明文选取了普朗特的就职演说，得到明文密文各个字符的频率如下：可以看出，在明文中，出现频率最高的是e:0.123 , t:0.089, a:0.082, o:0.081, n:0.070 在密文中，出现频率最高的是O:0.123, h:0.089, c:0.082, s:0.081, p:0.070经过计算可以知道字符o h c s p正是字符 e t a o n的加密变换由此可知，单表仿射加密是一种线性的加密，加密后字符的统计特性没有发生变化，只是字符不同而已。实验项目名 称实验二、序列密码成绩一、实验目的通过实现简单的线性反馈移位寄存器（LFSR），理解LFSR的工作原理、本原多项式的重要意

24、义。3、 实验内容1）利用CC+语言实现给定的LFSR；2）通过不同初始状态生成相应的序列，并观察它们的周期有什么特点；3）利用生成的序列对文本进行加/解密（按对应位作模2加运算）。给定的LFSR为：4、 实验步骤1） 在main函数中构建框架，分为三部分：初始化，打印序列，加密文本2） 初始化lfsr，这里用一个长度为5的整型数组来模拟移位寄存器，初始化时也只需要将这五个数组元素初始化即可。3） 打印序列，从移位寄存器的图示可以得知反馈函数为f = a4+a1;程序执行时先将a4+a1的结果保存，然后顺次移位a1<-a2,a2<-a3,a3<-a4,a4<-a4+a1

25、.因为lfsr一定具有周期性，且周期最大为2n-1,所以这里我们连续输出序列长度为2n+5,由此来观察周期性4） 在对文本进行加密时，每次从文件中读取一个数字0或者1，然后用lfsr顺次产生一个元素，进行模二加得到密文源代码：#include <stdio.h>#include<stdlib.h>#include<math.h>#define NUM 5/* run this program using the console pauser or add your own getch, system("pause") or input l

26、oop */int main() void init_lfsr(int *); void print_sequence(int *); void encrypt(int *); int regNUM ; int choice ; while(1) printf("nplease input your choice:n"); printf("t1.init lfsr 2.print sequence 3.encrpt text 4.exitn"); scanf("%d",&choice); getchar(); switch(c

27、hoice) case 1: init_lfsr(reg); break; case 2: print_sequence(reg); break; case 3: encrypt(reg); break; case 4: exit(2); return 0;void init_lfsr(int *reg) void print_sequence(int *r); printf("please input initial sequence:n"); for(int i=0; i<NUM; i+) scanf("%d",&regi); regi

28、%=2; getchar(); /*for(int i=0; i<2; i+) for(int j=0; j<2; j+) for(int k=0; k<2; k+) for(int m=0; m<2; m+) for(int n=0; n<2; n+) reg0=i;reg1=j;reg2=k;reg3=m,reg4=n; printf("ninitial sequence is :n"); for(int i=0; i<NUM; i+) printf("%d",regi); /getchar(); printf(&

29、quot;n"); print_sequence(reg); */ printf("initial sequence is :n"); for(int i=0; i<NUM; i+) printf("%d",regi); /getchar(); printf("n");void print_sequence(int *r) int temp ; int count = 0 ; int change_row = 0; int *reg = (int *)malloc(NUM*sizeof(int); if(reg = N

30、ULL) exit(4); for(int i=0; i<NUM; i+) regi = ri; while(count <= pow(2,NUM)+5) printf("%d ",reg0); change_row + ; if(change_row % 20 = 0) printf("n") ; temp = (reg0+reg3)%2; int i; for( i=0; i<NUM-1; i+) regi = regi+1 ; regi = temp ; count+ ; free(reg); void encrypt(int *

31、r) FILE *fp1,*fp2 ; int plain ; int cypher ; int temp ; int *reg = (int *)malloc(NUM*sizeof(int); if(reg = NULL) exit(4); for(int i=0; i<NUM; i+) regi = ri; if(fp1 = fopen("plain.txt","r")=NULL) printf("can't open file plian.txt!"); exit(0); if(fp2 = fopen("

32、cypher.txt","a")=NULL) printf("can't open file cypher.txt!"); exit(1); / fseek(fp1,0L,SEEK_SET); / fscanf(fp1,"%d",&plain); char ch1; ch0 = fgetc(fp1);/ printf("char:%cn",ch0); plain = atoi(ch);/ printf("plain:%dn",plain); while(!feof(fp

33、1) cypher = (plain + reg0)%2 ; fprintf(fp2,"%d",cypher); printf("%d",cypher); temp = (reg0+reg3)%2; int i; for( i=0; i<NUM-1; i+) regi = regi+1 ; regi = temp ; /fscanf(fp1,"%d",&plain); ch0 = fgetc(fp1);/ printf("char:%cn",ch0); plain = atoi(ch);/ print

34、f("plain:%dn",plain); fclose(fp1); fclose(fp2); free(reg); 四、实验结果及分析对源程序稍作修改，可以得到每个初始序列对应生成序列，如下：可以看出，除去第一个初始序列为全0的，其他31个序列的周期都是31，也即lfsr的最大周期25-1对文本进行加密，首先从文本plain.txt中读取明文，然后利用之前初始化得到的序列对明文进行加密，将结果写入cypher.txt中，同时打印输出到控制台，结果如下：实验项目名 称实验三、DES算法的实现成绩一、实验目的通过实现DES/AES算法，加深对DES/AES算法的理解，同时学习

35、组合密码常用的代换、移位等运算的实现。4、 实验内容1）利用CC+实现DES/AES算法的加、解密运算。5、 实验步骤（1） 初始置换，用初始置换IP表对明文进行置换（2） 密钥置换，除去每个字节的第八位，剩下56位密钥，然后在DES的每一轮中，从56位密钥中选出48位（3） 扩展置换，将数据的右半部份从32位扩展到48位（4） S盒代替，将上述的48位输入再次转换位32位输出（5） 逆初始置换源代码：#include <stdio.h>#include "time.h"#include "stdlib.h"#include "st

36、ring.h" static int IP_Table64 = 57,49,41,33,25,17,9,1, 59,51,43,35,27,19,11,3, 61,53,45,37,29,21,13,5, 63,55,47,39,31,23,15,7, 56,48,40,32,24,16,8,0, 58,50,42,34,26,18,10,2, 60,52,44,36,28,20,12,4, 62,54,46,38,30,22,14,6; static int IP_1_Table64 = 39,7,47,15,55,23,63,31, 38,6,46,14,54,22,62,30,

37、 37,5,45,13,53,21,61,29, 36,4,44,12,52,20,60,28, 35,3,43,11,51,19,59,27, 34,2,42,10,50,18,58,26, 33,1,41,9,49,17,57,25, 32,0,40,8,48,16,56,24;static int E_Table48 = 31, 0, 1, 2, 3, 4, 3, 4, 5, 6, 7, 8, 7, 8,9,10,11,12, 11,12,13,14,15,16, 15,16,17,18,19,20, 19,20,21,22,23,24, 23,24,25,26,27,28, 27,28

38、,29,30,31, 0;static int P_Table32 = 15,6,19,20,28,11,27,16, 0,14,22,25,4,17,30,9, 1,7,23,13,31,26,2,8, 18,12,29,5,21,10,3,24;static int S8416 =14,4,13,1,2,15,11,8,3,10,6,12,5,9,0,7,0,15,7,4,14,2,13,1,10,6,12,11,9,5,3,8,4,1,14,8,13,6,2,11,15,12,9,7,3,10,5,0,15,12,8,2,4,9,1,7,5,11,3,14,10,0,6,13, 15,1

39、,8,14,6,11,3,4,9,7,2,13,12,0,5,10,3,13,4,7,15,2,8,14,12,0,1,10,6,9,11,5,0,14,7,11,10,4,13,1,5,8,12,6,9,3,2,15,13,8,10,1,3,15,4,2,11,6,7,12,0,5,14,9, 10,0,9,14,6,3,15,5,1,13,12,7,11,4,2,8,13,7,0,9,3,4,6,10,2,8,5,14,12,11,15,1,13,6,4,9,8,15,3,0,11,1,2,12,5,10,14,7,1,10,13,0,6,9,8,7,4,15,14,3,11,5,2,12

40、, 7,13,14,3,0,6,9,10,1,2,8,5,11,12,4,15,13,8,11,5,6,15,0,3,4,7,2,12,1,10,14,9,10,6,9,0,12,11,7,13,15,1,3,14,5,2,8,4,3,15,0,6,10,1,13,8,9,4,5,11,12,7,2,14, 2,12,4,1,7,10,11,6,8,5,3,15,13,0,14,9,14,11,2,12,4,7,13,1,5,0,15,10,3,9,8,6,4,2,1,11,10,13,7,8,15,9,12,5,6,3,0,14,11,8,12,7,1,14,2,13,6,15,0,9,10

41、,4,5,3, 12,1,10,15,9,2,6,8,0,13,3,4,14,7,5,11,10,15,4,2,7,12,9,5,6,1,13,14,0,11,3,8,9,14,15,5,2,8,12,3,7,0,4,10,1,13,11,6,4,3,2,12,9,5,15,10,11,14,1,7,6,0,8,13, 4,11,2,14,15,0,8,13,3,12,9,7,5,10,6,1,13,0,11,7,4,9,1,10,14,3,5,12,2,15,8,6,1,4,11,13,12,3,7,14,10,15,6,8,0,5,9,2,6,11,13,8,1,4,10,7,9,5,0,

42、15,14,2,3,12, 13,2,8,4,6,15,11,1,10,9,3,14,5,0,12,7,1,15,13,8,10,3,7,4,12,5,6,11,0,14,9,2,7,11,4,1,9,12,14,2,0,6,10,13,15,3,5,8,2,1,14,7,4,10,8,13,15,12,9,0,3,5,6,11; static int PC_156 = 56,48,40,32,24,16,8, 0,57,49,41,33,25,17, 9,1,58,50,42,34,26, 18,10,2,59,51,43,35, 62,54,46,38,30,22,14, 6,61,53,

43、45,37,29,21, 13,5,60,52,44,36,28, 20,12,4,27,19,11,3;static int PC_248 = 13,16,10,23,0,4,2,27, 14,5,20,9,22,18,11,3, 25,7,15,6,26,19,12,1, 40,51,30,36,46,54,29,39, 50,44,32,46,43,48,38,55, 33,52,45,41,49,35,28,31;static int MOVE_TIMES16 = 1,1,2,2,2,2,2,2,1,2,2,2,2,2,2,1;static unsigned char SubKey16

44、48;void ByteToBit(unsigned char *ch,unsigned char *bit);void BitToByte(unsigned char *bit,unsigned char *ch);void Xor(unsigned char *A,unsigned char *B,int len);void Transform(unsigned char *str,unsigned char *change,int *table,int len);void Rotatel(unsigned char *str, int loop);void MakeSubKeys(uns

45、igned char *k);void F_fun(unsigned char *M,unsigned char *key);void DES_Encrypt(unsigned char *plain,unsigned char *cipher);void DES_Decrypt(unsigned char *cipher,unsigned char *plain);void ByteToBit(unsigned char* ch,unsigned char* bit) int i,j; for(i=0; i<8; i+) for(j = 0;j< 8; j+) biti*8+j = (*(ch+i)>>j)&1; void BitToByte(unsigned char *bit, unsigned char *ch) int i,j; memset(ch,0,9); for(i=0; i<8; i+) for(j = 0;j< 8; j+) chi |= *(bit + j+8*i)<<j; void Transform(unsigned char *str,unsigned