SUSCTF-2022 部分WriteUp

import mathe = [207, 359, 220, 224, 352, 315, 359, 374, 290, 310, 277, 507, 391, 513, 423, 392, 508, 383, 440, 322, 420, 427, 503, 460, 295, 318, 245, 302, 407, 414, 410, 130, 369, 317]a = 'AnEWmuLTiPLyis_etimes_wiLLbEcomE_B'#b = ['01001', '11010', '00110', '11000', '00001', '10000', '10001', '01011', '01011', '01011', '01110', '00110', '00101', '10010', '00100', '10100', '11010', '10110', '11000', '00101', '11000', '01001', '10010', '01010', '01100', '11010', '10000', '10001', '00101', '00010', '11001', '01000', '01000', '00011']e10= []alpha = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'b = 'NQHFEAOUUUSHLMCJQRFLFNMKGQAOLDWBBI'
result = [9,26,6,]#207*26+1/65#101*95 = 26*369+1
for i in range(34):  #rint(95*101/e[i])
  a1 = ord(a[i])  #print('a：',a1)  n = e[i]  flag = chr(round((n * 26 + 1)/a1))  print(flag)'''for i in e:  e10.append(bin(i)[2:])
print(e10)  '''

#include <stdio.h>#include <stdint.h>// unsigned char ida_chars[] =//{// 0x88, 0x34, 0xD9, 0x48, 0x4C, 0x14, 0x0C, 0x03, 0xC2, 0x78,//  0xEB, 0x52, 0xED, 0xE5, 0x9C, 0xED,0xE6, 0xED, 0x1F, 0xAE, 0x6D, 0x12, 0x5A, 0xBA, 0xAA, //0x84,//  0x92, 0xCF, 0xE3, 0xF2, 0xE0, 0x65//};//解密函数// void decrypt (uint32_t* v, uint32_t* k) {//     uint32_t v0=v[0], v1=v[1], sum=0xC6EF3720, i;  /* set up *///     uint32_t delta=0x9e3779b9;                     /* a key schedule constant *///     uint32_t k0=k[0], k1=k[1], k2=k[2], k3=k[3];   /* cache key *///     for (i=0; i<32; i++) {                         /* basic cycle start *///         v1 -= ((v0<<4) + k2) ^ (v0 + sum) ^ ((v0>>5) + k3);//         v0 -= ((v1<<4) + k0) ^ (v1 + sum) ^ ((v1>>5) + k1);//         sum -= delta;//     }                                              /* end cycle *///     v[0]=v0; v[1]=v1;// }void decrypt (uint32_t* v, uint32_t* k) {    uint32_t v0=v[0], v1=v[1], sum=0x9e3779b9*32, i;  /* set up */    uint32_t delta=0x9e3779b9;                        /* a key schedule constant */    uint32_t k0=k[0], k1=k[1], k2=k[2], k3=k[3];/* cache key */    for (i=0; i<32; i++) {                         /* basic cycle start */         v1 -= ((v0<<4) + k2) ^ (v0 + sum) ^ ((v0>>5) + k3);        v0 -= ((v1<<4) + k0) ^ (v1 + sum) ^ ((v1>>5) + k1);        sum -= delta;    }                                              /* end cycle */    v[0]=v0; v[1]=v1;}int main(){    //unsigned char key[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=";    uint32_t v[6]={0x3e8947cb,0xcc944639,0x31358388,0x3b0b6893,0xda627361,0x3b2e6427} ,k[4]={17477,0x4144,0x4245,0x4546};    //uint32_t *k = (uint32_t*)key;    //uint32_t *v = (uint32_t*)ida_chars;    // v为要加密的数据是两个32位无符号整数    // k为加密解密密钥，为4个32位无符号整数，即密钥长度为128位    printf("n");    decrypt(v, k);    decrypt(v+2, k);     decrypt(v+4, k);    // decrypt(v+6, k);     //decrypt(v+8, k);     //decrypt(v+10, k);     //decrypt(v+12, k);        //decrypt(v+14, k);        //decrypt(v+16, k);        //decrypt(v+18, k);       // decrypt(v+20, k);        //decrypt(v+22, k);        //decrypt(v+24, k);    for(int i=0;i<6;i++){      printf("%x ",v[i]);    }    printf("n");    //for(int i=0;i<32;i++){    //    printf("%c",ida_chars[i]);    //}    return 0;}

a=[0x58,0x42,0x76,0x66,0x61,0x45,0x64,0x51,0x76,0x62,0x63,0x72,0x78,0x50,0x42,0x68,0x38,0x41,0x4f,0x63,0x4a,0x36,0x67,0x41]for i in range(len(a)):     print(chr(a[i]),end='')