该文讨论了一种RSA加密的实现过程涉及到数据(data)的指数运算(e02^16)以及模N的计算。通过检查(data**e0-1)与N的最大公约数(GCD)尝试找到N的因子p。计算得出p后求得q进而确定欧拉函数φ并计算私钥d最后解密得到原始信息。(data**e-data)modN0e0e-1;(data**e0-1)modN0e0为2**16因式分解 data**(2**i)-1,可能存在p或者q的因子尝试与N进行GCDdata0x372f0e88f6f7189da7c06ed49e87e0664b988ecbee583586dfd1c6af99bf20345ae7442012c6807b3493d8936f5b48e553f614754deb3da6230fa1e16a8d5953a94c886699fc2bf409556264d5dced76a1780a90fd22f3701fdbcb183ddab4046affdc4dc6379090f79f4cd50673b24d0b08458cdbe509d60a4ad88a7b4e2921 e0x10001 e0e-1 N 0x7fe8cafec59886e9318830f33747cafd200588406e7c42741859e15994ab62410438991ab5d9fc94f386219e3c27d6ffc73754f791e7b2c565611f8fe5054dd132b8c4f3eadcf1180cd8f2a3cc756b06996f2d5b67c390adcba9d444697b13d12b2badfc3c7d5459df16a047ca25f4d18570cd6fa727aed46394576cfdb56b41 e 0x10001 c 0x5233da71cc1dc1c5f21039f51eb51c80657e1af217d563aa25a8104a4e84a42379040ecdfdd5afa191156ccb40b6f188f4ad96c58922428c4c0bc17fd5384456853e139afde40c3f95988879629297f48d0efa6b335716a4c24bfee36f714d34a4e810a9689e93a0af8502528844ae578100b0188a2790518c695c095c9d677b while e01: if gcd(pow(data,e0,N)-1,N)1 and gcd(pow(data,e0,N)-1,N)N: pgcd(data**e0-1,N) break e0e0//2 print(p) qN//p assert(p*qN) phi (p - 1) * (q - 1) d invert(e,phi) mpow(c,d,N) print(long_to_bytes(m))