"Karioj" Discrete math Spring semester programming Test 1

A. Caesar Password Test Instructions:

Give you k1,k2, and a string of plaintext, a bunch of ciphertext.

PlainText is encrypted with K1, ciphertext is decrypted with K2.

For clear text to convert the letters into uppercase letters, all non-letters are deleted.

Amount: To take into account that the modulus may become negative, so add 26 and then modulo. Code:

#include <stdio.h> #define N 85int K1, K2;char plain[n], cipher[n];void init (char s[])//preprocessing, converted to uppercase {    int i;    for (i = 0; s[i]; i++)        if (S[i] >= ' A ' && s[i] <= ' z ')            s[i] = s[i]-' a ' + ' a ';} void Encrypt (char s[],int k,int f)//Plus decryption {    init (s);    int i;    for (i = 0; s[i]; i++)        if (s[i] >= ' A ' && s[i] <= ' Z ')            printf ("%c", ((S[i]-' A ' + k*f + +)%  + ' A '));    printf ("\ n");} int main () {    scanf ("%d,%d", &k1, &K2);    Gets (plain);    Gets (cipher);    Encrypt (plain,k1,1);    Encrypt (cipher,k2,-1);    return 0;}

B. RSA Encryption Test Instructions:

Give you n,e, and a bunch of clear text. encrypt the plaintext with (n,e). Converts the plaintext letter to a number, divided by 8 digits, and the less part complements 0. There are non-letter deletions in the plaintext, a and a turn into numbers, and Z and Z turn into numbers of 25. Clear text number 8 bits of each paragraph corresponding to the ciphertext is also required to be 8 bits, if less than 8 bits, the front top up 0.

For clear text to convert the letters into uppercase letters, all non-letters are deleted.

Add:

RSA encryption is the conversion of letters to two-digit, segmented processing, such as every eight paragraphs, M is the clear-Text Number field, C for the ciphertext field, c=me%n.

Code:

#include <stdio.h> #include <string.h> #define N 1000#define ll long longll N, E;char plain[n];ll Qpow (ll A, LL b)    Fast Power {ll k = a% n;    ll ans = 1;        while (b) {if (b & 1) ans = (ans * k)% n;        K = (k * k)% n;    b >>= 1; } return ans;    void Init (char s[])//preprocessing, converted to uppercase {int i; for (i = 0; s[i]; i++) if (S[i] >= ' A ' && s[i] <= ' z ') s[i] = s[i]-' a ' + ' a ';}    void Encrypt (char s[])//encryption {init (s);    int k = 0,i;    ll block = 0; for (i = 0; s[i]; i++) if (s[i] >= ' A ' && s[i] <= ' Z ')//If it is the letter {block = block * 100 + (S[i]-' A ');            Clear text corresponding to the number string k++;                if (k = = 4)//enough 8 digits {printf ("%08lld", QPOW (block, e)% 100000000);//output ciphertext, this module does not know whether it is necessary, the title does not say the upper limit of n                block = 0;            k = 0;          }} if (k)//The remaining plaintext is to be followed by 0 {while (k! = 4) {block = block * 100;  k++;    } printf ("%08lld", QPOW (block, e)% 100000000);    }}int Main () {//Freopen ("In.txt", "R", stdin);    scanf ("%lld%lld", &n, &e);    Gets (plain);    Encrypt (plain); return 0;}

C. RSA Decryption Test Instructions:

Give you n,e, and a bunch of clear text. Encrypt the plaintext with (n,e). Converts the plaintext letter to a number, divided by 8 digits, and the less part complements 0. There are non-letter deletions in the plaintext, a and a turn into numbers, and Z and Z turn into numbers of 25. Clear text number 8 bits of each paragraph corresponding to the ciphertext is also required to be 8 bits, if less than 8 bits, the front top up 0.

For clear text to convert the letters into uppercase letters, all non-letters are deleted.

Add:

RSA encryption is the conversion of letters to two-digit, segmented processing, such as every eight paragraphs, M is the clear-Text Number field, C for the ciphertext field, c=me%n.

Code:

#include <stdio.h> #include <string.h> #define N 802#define ll long longll N, E, p, Q, D, X, Y;char cipher[n <& Lt        1], Plain[10];ll EXGCD (ll A, ll b)//Expand Euclid to seek the inverse of the meta {if (b = = 0) {x = 1;        y = 0;    return A;    } ll R = EXGCD (b, a% B);    ll tmp = x;    x = y;    y = tmp-a/b * y; return r;}    ll Qpow (ll A, ll b)//fast Power {ll k = a% n;    ll ans = 1;        while (b) {if (b & 1) ans = (ans * k)% n;        K = (k * k)% n;    b >>= 1; } return ans;    void init ()//ask P and Q, and d{int i;            for (i = 2; i < n; i++) if (n% i = = 0) {p = i;            Q = n/i;        Break    } ll M;    M = (p-1) * (q-1);    EXGCD (E, M);    d = (x% m + m)% m;//the inverse of e d}void decrypt ()//decryption {LL block = 0;    int k = 0, I, J;        for (i = 0; cipher[i]; i++) {block = block * + cipher[i]-' 0 ';//ciphertext is removed from the string to k++; if (k = = 8)//ciphertext reaches 8 digits {memset (plain, 0, sizEOF plain);//Clear Clear text string block = QPOW (block, d);//calculation of the plaintext number string for (j = 3; J >= 0; j--)//Calculate two digits each time and there is a clear text string, Because it is taken from the back, it is stored upside down {plain[j] = block% + ' A ';//take the back two digits block/= 100;//Remove the last two numbers Word} printf ("%s", plain);//output plaintext K = 0;//Empty counter}}}int main () {//Freopen ("in    . txt "," R ", stdin);    scanf ("%lld%lld", &n, &e);    Gets (cipher);    Init ();    printf ("%d\n", D);    Decrypt (); return 0;}

"Karioj" Discrete math Spring semester programming Test 1