From b3ce1dfaef6b549115017ef2f637f7813a2bd6e4 Mon Sep 17 00:00:00 2001
From: sangge <2251250136@qq.com>
Date: Mon, 8 Sep 2025 17:25:30 +0800
Subject: [PATCH] feat: finish p16

---
 cryptopal_book/src/challenge_16.md |  2 +-
 problems/p16/Cargo.toml            |  2 ++
 problems/p16/src/main.rs           | 49 ++++++++++++++++++++++++++++--
 3 files changed, 50 insertions(+), 3 deletions(-)

diff --git a/cryptopal_book/src/challenge_16.md b/cryptopal_book/src/challenge_16.md
index 6571f74..9cc80f9 100644
--- a/cryptopal_book/src/challenge_16.md
+++ b/cryptopal_book/src/challenge_16.md
@@ -26,7 +26,7 @@
 
 如果你正确编写了第一个函数，应该*不*可能向其提供会生成第二个函数正在寻找的字符串的用户输入。我们必须破解密码才能做到这一点。
 
-相反，修改密文（不知道AES密钥）来完成这个目标。
+相反，修改**密文**（不知道AES密钥）来完成这个目标。
 
 你依赖的事实是，在CBC模式中，密文块中的1位错误：
 
diff --git a/problems/p16/Cargo.toml b/problems/p16/Cargo.toml
index ed22fc5..b195a35 100644
--- a/problems/p16/Cargo.toml
+++ b/problems/p16/Cargo.toml
@@ -4,3 +4,5 @@ version = "0.1.0"
 edition = "2024"
 
 [dependencies]
+common = { path = "../../common/" }
+rand = { workspace = true }
diff --git a/problems/p16/src/main.rs b/problems/p16/src/main.rs
index e7a11a9..313f442 100644
--- a/problems/p16/src/main.rs
+++ b/problems/p16/src/main.rs
@@ -1,3 +1,48 @@
-fn main() {
-    println!("Hello, world!");
+use common::{aes_cbc_dec, aes_cbc_enc, pkcs7_padding, pkcs7_unpadding};
+use rand::{Rng, rng};
+fn append_and_enc(input: String, key: &[u8; 16]) -> Vec<u8> {
+    let input = input.replace(";", "\\;").replace("=", "\\=");
+    let prefix = "comment1=cooking%20MCs;userdata=".to_string();
+    let suffix = ";comment2=%20like%20a%20pound%20of%20bacon";
+    // println!("{}, {}", prefix.len(), suffix.len()); // 32, 42
+
+    let mut buffer = prefix + &input + suffix;
+    let input = unsafe { buffer.as_mut_vec() };
+    pkcs7_padding(input, 16);
+
+    let iv = [0u8; 16];
+
+    aes_cbc_enc(input, key, &iv).unwrap()
+}
+
+fn verify(cipher: &[u8], key: &[u8; 16]) -> bool {
+    let iv = [0u8; 16];
+    let plaintext = aes_cbc_dec(cipher, key, &iv).unwrap();
+    let plaintext = pkcs7_unpadding(&plaintext).unwrap();
+    let plaintext = String::from_utf8_lossy(&plaintext);
+    println!("{}", plaintext);
+    plaintext.contains(";admin=true;")
+}
+
+fn main() {
+    let mut rng = rng();
+    let key: [u8; 16] = rng.random();
+
+    // 构造输入，第一个块用于反转破坏
+    let input = "A".repeat(16) + &"A".repeat(5) + "XadminXtrue";
+    let mut cipher = append_and_enc(input, &key);
+
+    // 计算目标位置：prefix(32) + padding(16) + target_start = 48 + 1 = 49
+    // 目标在第4块，需要修改第3块
+    let block_to_modify = 2 * 16; // 第3块起始位置
+
+    // 修改密文来翻转目标位
+    cipher[block_to_modify + 5] ^= b'X' ^ b';'; // X -> ;
+    cipher[block_to_modify + 11] ^= b'X' ^ b'='; // X -> =
+
+    if verify(&cipher, &key) {
+        println!("Attack successful!");
+    } else {
+        println!("Attack failed");
+    }
 }