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翻斗

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简单的循环加解密算法

翻斗
2015-11-02 / 0 评论 / 0 点赞 / 969 阅读 / 1,290 字

有时候需要自己弄个简单的加密解密算法。

#include<iostream>
#include<stdio.h>
using namespace std;
const char* encrypt(const char* ip, unsigned int key[], int len)//加密函数
{  
    //要保证(x + n)% 126 + 32 < 126 所以要保证n < 94
    //也就是key数组里面必须为小于94的数
    //unsigned int b[]={2,7,18,23,91,5,4};
    int i=0;
    char* dest = (char*)calloc(strlen(ip), sizeof(char));

    while(*ip)
    {

        int x = (int)(*ip) + key[i % len];
        if(x>126){
            x = x % 126 + 32;
        }
        *dest = x;
        i++,ip++,dest++;
    }        
    dest -= i;
    ip -= i;
    printf("[%s=>%s]\n", ip, dest);
    return dest;
}

const char* decrypt(const char* src, unsigned int key[], int len){
    char* dest = strdup(src); 
    //int b[]={2,7,18,23,91,5,4};
    int i=0;
    while(*dest)
    {

        int x = (int)*dest - key[i % len];
        if( x < 32){
            x = x - 32 + 126;
        }
        *dest = x;
        i++,dest++;
    }
    dest -= i;
    return dest;
}

int main(){
unsigned int b[]={2,7,18,23,91,5,4};
const char* ss = encrypt("/libdje1.so", b, 7);
printf("%s\n", decrypt(ss, b, 7));
const char * e100 = encrypt("100", b, 7);
const char* e101 = encrypt("/sys/devices/system/cpu/cpu0/cpufreq/scaling_available_frequencies", b, 7);
printf("%s\n", e101);
const char* xx = "1z-,,iixpu|p4w{z(|j4gr|Azmz41j$.cwis6'z^qmpnqxsfmnht%bdjtl%.bsgkl'";
printf("%s\n", decrypt(xx, b, 7));
printf("%s\n", encrypt("&cpu_freq=", b, 7));
const char* cpu_freq = "(j$.\\kvgxO";
printf("%s\n", decrypt(cpu_freq, b, 7));

return 1;
}

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