From cd8379ae5ed55aa1c0a6ab44d11a61cb285435de Mon Sep 17 00:00:00 2001
From: mwatson-aprilu <xkeuzdzlyz4822@hotmail.com>
Date: Sun, 26 Apr 2026 19:43:04 +0800
Subject: [PATCH] =?UTF-8?q?Create=2025-Cry-shuiyue-=E6=9D=A8=E6=99=B6.md?=
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Added detailed explanation and code for the third problem involving RSA encryption and decryption using Python.
---
 ...5-Cry-shuiyue-\346\235\250\346\231\266.md" | 57 ++++++++++++++++++-
 1 file changed, 56 insertions(+), 1 deletion(-)

diff --git "a/Cry/submissions/Cry/submissions/25-Cry-shuiyue-\346\235\250\346\231\266.md" "b/Cry/submissions/Cry/submissions/25-Cry-shuiyue-\346\235\250\346\231\266.md"
index 3ff8f72..af358d1 100644
--- "a/Cry/submissions/Cry/submissions/25-Cry-shuiyue-\346\235\250\346\231\266.md"
+++ "b/Cry/submissions/Cry/submissions/25-Cry-shuiyue-\346\235\250\346\231\266.md"
@@ -93,7 +93,62 @@ c = 4944265592362530990938501754537183733907901996253368795483052159403178819389
 2.知识补充（自我提醒）：     
 >题目中的dp,dq是由p,q分别减一再取模得到，可以简化计算。      
 3.由于dp,dq可看作密钥的一部分，故而先'pow'算出dp,dq。    
-4.利用中国剩余定理得到m，再将m转化为字符型，得到flag。   
+4.利用中国剩余定理得到m，再将m转化为字符型，得到flag。
+## 第三题 three
+
+```py
+import libnum
+import gmpy2
+import uuid
+flag = "flag{" + str(uuid.uuid4()) + "}"
+m = libnum.s2n(flag)
+print(gmpy2.bit_length(m ** 3))
+while True:
+    p = libnum.generate_prime(504)
+    q = libnum.generate_prime(504)
+    n = p * q
+    phi_n = (p - 1) * (q - 1)
+    e = 3
+    if gmpy2.gcd(e, phi_n) == 1 and  phi_n%e !=0:
+        break
+
+c = pow(m, e, n)
+print("n=", n)
+print("e=", e)
+print("c=", c)
+'''
+n= 1429271935073130420990643850236689595291596400638807867713275629712107176185520444788761901754075622136449476721042807647099083589759664014467696405646188198112649666177678289522724798981531832903351705482763333352669597586657968404428319154104297454666647063679624028208478294081894177792365782518110223
+e= 3
+c= 175676150266632261170048224865383344382701466592480820852821987310883907646453945269873798553604154950679789706816103734968119191264040967221804258086782193045472222806462922151478555259607753533988574525462498108797766467552606740700525351657072574897006690469089852849417334378963664086027896356417125
+'''
+```
+
+思路思路：        
+1.题目里 e=3，c = m³ mod n           
+2.因为明文 m 是 flag+uuid，长度很短，m³ 直接小于模数 n，没有发生取模运算       
+3.所以：c = m³，直接对 c 开三次方就能得到 m    
+4.再把数字 m 转回字符串就是 flag    
+
+```py   
+from Crypto.Util.number import *
+import gmpy2
+n= 1429271935073130420990643850236689595291596400638807867713275629712107176185520444788761901754075622136449476721042807647099083589759664014467696405646188198112649666177678289522724798981531832903351705482763333352669597586657968404428319154104297454666647063679624028208478294081894177792365782518110223
+e= 3
+c= 175676150266632261170048224865383344382701466592480820852821987310883907646453945269873798553604154950679789706816103734968119191264040967221804258086782193045472222806462922151478555259607753533988574525462498108797766467552606740700525351657072574897006690469089852849417334378963664086027896356417125
+h = pow(2, 1005)
+if (h < n):
+    print("YES")
+m = int(gmpy2.iroot(c, 3)[0])
+flag = libnum.n2s(m)
+print(flag)     
+```
+
+最后输出：
+
+```txt
+YES
+b'flag{d07cc7cf-9d78-4bfb-9449-d099b0329563}'
+```  
 ##为什么要选择这个方向：  
 >实践出真知