RSA算法代码#inclede<stdio.h> #include<math.h> int Prime(int _m){int i,k; if(m<2) return 0;补充剩余代码

int Prime(int _m){
int i,k;
if(m<2) return 0;
k = sqrt(m);
for(i=2;i<=k;i++){
if(m%i==0) return 0;
}
return 1;
}
// RSA加密
int RSA_Encrypt(int p,int q,int e,int m){
int n = p*q;
int fn = (p-1)*(q-1);
int i,d;
for(i=1;i<fn;i++){
if((i*e)%fn==1){
d = i;
break;
}
}
int c = pow(m,e);
c = fmod(c,n);
return c;
}
// RSA解密
int RSA_Decrypt(int p,int q,int e,int c){
int n = p*q;
int fn = (p-1)*(q-1);
int i,d;
for(i=1;i<fn;i++){
if((i*e)%fn==1){
d = i;
break;
}
}
int m = pow(c,d);
m = fmod(m,n);
return m;
}

相关问题

C语言实现RSA算法代码

以下是C语言实现RSA算法的代码示例：

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int main() {
    int p, q, n, phi, e, d, plain_text, cipher_text, decrypted_text;
    printf("Enter two prime numbers p and q: ");
    scanf("%d %d", &p, &q);

    n = p * q;
    phi = (p - 1) * (q - 1);

    // Select an integer e such that 1 < e < phi and gcd(e, phi) = 1.
    printf("Enter the value of e such that gcd(e, phi) = 1: ");
    scanf("%d", &e);

    // Calculate d such that d ≡ e^(-1) (mod phi).
    d = fmod(pow(e, -1), phi);

    printf("Enter the plain text: ");
    scanf("%d", &plain_text);

    // Encryption: c = (plain_text^e) % n.
    cipher_text = fmod(pow(plain_text, e), n);
    printf("Cipher text: %d\n", cipher_text);

    // Decryption: plain_text = (cipher_text^d) % n.
    decrypted_text = fmod(pow(cipher_text, d), n);
    printf("Decrypted text: %d\n", decrypted_text);

    return 0;
}

在该示例中，用户需要输入两个素数`p`和`q`、加密指数`e`以及明文`plain_text`。程序计算出了`n`和`phi`，然后计算出了私钥`d`。使用公钥`(n, e)`对明文进行加密，使用私钥`(n, d)`对密文进行解密。

请注意，此示例仅用于演示目的，实际使用中需要考虑更多的安全性和性能方面的问题。

RSA算法的加解密C语言代码

以下是RSA算法的加解密C语言代码：

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>

long int p, q, n, t, flag, e[100], d[100], temp[100], j, m[100], en[100], i;
char msg[100];

int prime(long int);
void ce();
long int cd(long int);
void encrypt();
void decrypt();

int main() {
    printf("\nEnter two prime numbers: ");
    scanf("%ld %ld", &p, &q);

    flag = prime(p);
    if(flag == 0) {
        printf("\n%d is not prime\n", p);
        exit(1);
    }

    flag = prime(q);
    if(flag == 0 || p == q) {
        printf("\n%d is not prime\n", q);
        exit(1);
    }

    printf("\nEnter message: ");
    fflush(stdin);
    scanf("%s", msg);
    for(i = 0; msg[i] != '\0'; i++)
        m[i] = msg[i];
    n = p * q;
    t = (p - 1) * (q - 1);
    ce();
    printf("\nPossible values of e and d are:\n");
    for(i = 0; i < j - 1; i++)
        printf("\n%ld\t%ld", e[i], d[i]);

    encrypt();
    decrypt();
    return 0;
}

int prime(long int pr) {
    int i;
    j = sqrt(pr);
    for(i = 2; i <= j; i++) {
        if(pr % i == 0)
            return 0;
    }
    return 1;
}

void ce() {
    int k;
    k = 0;
    for(i = 2; i < t; i++) {
        if(t % i == 0)
            continue;
        flag = prime(i);
        if(flag == 1 && i != p && i != q) {
            e[k] = i;
            flag = cd(e[k]);
            if(flag > 0) {
                d[k] = flag;
                k++;
            }
            if(k == 99)
                break;
        }
    }
}

long int cd(long int x) {
    long int k = 1;
    while(1) {
        k = k + t;
        if(k % x == 0)
            return(k / x);
    }
}

void encrypt() {
    long int pt, ct, key = e[0], k, len;
    i = 0;
    len = strlen(msg);
    while(i != len) {
        pt = m[i];
        pt = pt - 96;
        k = 1;
        for(j = 0; j < key; j++) {
            k = k * pt;
            k = k % n;
        }
        temp[i] = k;
        ct = k + 96;
        en[i] = ct;
        i++;
    }
    en[i] = -1;
    printf("\nEncrypted message: ");
    for(i = 0; en[i] != -1; i++)
        printf("%c", en[i]);
}

void decrypt() {
    long int pt, ct, key = d[0], k;
    i = 0;
    while(en[i] != -1) {
        ct = temp[i];
        k = 1;
        for(j = 0; j < key; j++) {
            k = k * ct;
            k = k % n;
        }
        pt = k + 96;
        m[i] = pt;
        i++;
    }
    m[i] = -1;
    printf("\nDecrypted message: ");
    for(i = 0; m[i] != -1; i++)
        printf("%c", m[i]);
    printf("\n");
}