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[原创]2021 KCTF 秋季赛 第二题 迷失丛林
发表于: 2021-11-17 11:53 12210

[原创]2021 KCTF 秋季赛 第二题 迷失丛林

kkHAIKE 活跃值
10
2021-11-17 11:53
12210

windows 程序通过查找 GetDlgItemTextA 引用找到关键入口
流程如下：

通过 第二个算法 有对换的特点，和 原始tb 唯一数字的特点

可以看出 tb 应该要满足 256 数字唯一

 
package main
 
import (
    "fmt"
)
 
// tb 404000
var t2tb = []byte{
    0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0xA2, 0x9B,
    0xF4, 0xDF, 0xAC, 0x7C, 0xA1, 0xC6, 0x16, 0xD0, 0x0F, 0xDD,
    0xDC, 0x73, 0xC5, 0x6B, 0xD1, 0x96, 0x47, 0xC2, 0x26, 0x67,
    0x4E, 0x41, 0x82, 0x20, 0x56, 0x9A, 0x6E, 0x33, 0x92, 0x88,
    0x29, 0xB5, 0xB4, 0x71, 0xA9, 0xCE, 0xC3, 0x34, 0x50, 0x59,
    0xBF, 0x2D, 0x57, 0x22, 0xA6, 0x30, 0x04, 0xB2, 0xCD, 0x36,
    0xD5, 0x68, 0x4D, 0x5B, 0x45, 0x9E, 0x85, 0xCF, 0x9D, 0xCC,
    0x61, 0x78, 0x32, 0x76, 0x31, 0xE3, 0x80, 0xAD, 0x39, 0x4F,
    0xFA, 0x72, 0x83, 0x4C, 0x86, 0x60, 0xB7, 0xD7, 0x63, 0x0C,
    0x44, 0x35, 0xB3, 0x7B, 0x19, 0xD4, 0x69, 0x08, 0x0B, 0x1F,
    0x3D, 0x11, 0x79, 0xD3, 0xEE, 0x93, 0x42, 0xDE, 0x23, 0x3B,
    0x5D, 0x8D, 0xA5, 0x77, 0x5F, 0x58, 0xDB, 0x97, 0xF6, 0x7A,
    0x18, 0x52, 0x15, 0x74, 0x25, 0x62, 0x2C, 0x05, 0xE8, 0x0D,
    0x98, 0x2A, 0x43, 0xE2, 0xEF, 0x48, 0x87, 0x49, 0x1C, 0xCA,
    0x2B, 0xA7, 0x8A, 0x09, 0x81, 0xE7, 0x53, 0xAA, 0xFF, 0x6F,
    0x8E, 0x91, 0xF1, 0xF0, 0xA4, 0x46, 0x3A, 0x7D, 0x54, 0xEB,
    0x2F, 0xC1, 0xC0, 0x0E, 0xBD, 0xE1, 0x6C, 0x64, 0xBE, 0xE4,
    0x02, 0x3C, 0x5A, 0xA8, 0x9F, 0x37, 0xAF, 0xA0, 0x13, 0xED,
    0x1B, 0xEC, 0x8B, 0x3E, 0x7E, 0x27, 0x99, 0x75, 0xAB, 0xFE,
    0xD9, 0x3F, 0xF3, 0xEA, 0x70, 0xF7, 0x95, 0xBA, 0x1D, 0x40,
    0xB0, 0xF9, 0xE5, 0xF8, 0x06, 0xBC, 0xB6, 0x03, 0xC9, 0x10,
    0x9C, 0x2E, 0x89, 0x5C, 0x7F, 0xB1, 0x1A, 0xD6, 0x90, 0xAE,
    0xDA, 0xE6, 0x5E, 0xB9, 0x84, 0xE9, 0x55, 0xBB, 0xC7, 0x0A,
    0xE0, 0x66, 0xF2, 0xD8, 0xCB, 0x00, 0x12, 0xB8, 0x17, 0x94,
    0x6A, 0x4A, 0x01, 0x24, 0x14, 0x51, 0x07, 0x65, 0x21, 0xC8,
    0x38, 0xFD, 0x8F, 0xC4, 0xF5, 0xFC,
}
 
func getloss() (ret []byte) {
    m := make(map[byte]bool)
    for i := 8; i < 256; i++ {
        m[t2tb[i]] = true
    }
 
    for i := 0; i < 256; i++ {
        if !m[byte(i)] {
            ret = append(ret, byte(i))
        }
    }
    return
}
 
// 排列算法
func permutations(arr []byte) [][]byte {
    var helper func([]byte, int)
    res := [][]byte{}
 
    helper = func(arr []byte, n int) {
        if n == 1 {
            tmp := make([]byte, len(arr))
            copy(tmp, arr)
            res = append(res, tmp)
        } else {
            for i := 0; i < n; i++ {
                helper(arr, n-1)
                if n%2 == 1 {
                    tmp := arr[i]
                    arr[i] = arr[n-1]
                    arr[n-1] = tmp
                } else {
                    tmp := arr[0]
                    arr[0] = arr[n-1]
                    arr[n-1] = tmp
                }
            }
        }
    }
    helper(arr, len(arr))
    return res
}
 
// 第一个校验算法
func checksk(c [256]byte) bool {
    var sss [256 * 2]byte
    var kkk [256][256]byte
    maxn := []int{2, 4, 8, 0x10, 0x20, 0x40, 0x80}
    for i := 0; i < 256; i++ {
        sss[0] = c[i]
        sss[1] = byte(i + 1)
        var ps int
        k := maxn[0]
        for _, n := range maxn {
            for j := 0; j < n; j++ {
                sss[k] = c[sss[ps]]
                sss[k+1] = sss[ps] + 1
                k += 2
                ps++
            }
        }
 
        for j := 0; j < 256; j++ {
            kkk[i][sss[ps]]++
            ps++
        }
    }
 
    var r1, r2, r3, r4 int
    for i := 0; i < 256; i++ {
        if kkk[i][0] != 0 {
            r1++
        }
        if kkk[i][40-26] != 0 {
            r2++
        }
        if kkk[i][40] != 0 {
            r3++
        }
        if kkk[i][40+39] != 0 {
            r4++
        }
    }
 
    return r1 == 169 && r2 == 172 && r3 == 167 && r4 > 200
}
 
// 要反一下打印
func printHex(p []byte) {
    for _, v := range p {
        c := ((v & 0xf) << 4) | ((v & 0xf0) >> 4)
        fmt.Printf("%X", c)
    }
    fmt.Println()
}
 
// 求解第一部分
func t2p1() {
    // 获取 tb 中缺失的 8 个数
    loss := getloss()
    // 排列所有可能 8! = 40320
    for _, pad := range permutations(loss) {
        var t [256]byte
        copy(t[:8], pad)
        copy(t[8:], t2tb[8:])
        // 求出满足第一个校验的 pad
        if checksk(t) {
            printHex(pad)
        }
    }
}
 
// 输入 前半部分 结果后
// 直接 dump 内存获取 tb变换 的结果
var t2tb2 = []byte{
    0xC1, 0x9B, 0x7F, 0x58, 0x64, 0xD5, 0x77, 0x21, 0x74, 0xEB, 0x14, 0xBF, 0xDF, 0x25, 0x5A, 0x37,
    0x85, 0x2C, 0xAF, 0x8C, 0xDA, 0x26, 0xE2, 0x7A, 0x87, 0x4C, 0x60, 0x99, 0x54, 0x3C, 0x95, 0xC0,
    0xB9, 0x0C, 0xBC, 0x0E, 0xE7, 0x2D, 0x86, 0xBE, 0x67, 0xD3, 0xD8, 0xFC, 0x30, 0xB6, 0xC8, 0x57,
    0x1E, 0x62, 0x3E, 0xCE, 0xA0, 0xCD, 0xF5, 0xEE, 0xA7, 0xCF, 0x45, 0xFE, 0xD0, 0x80, 0x05, 0xAD,
    0x13, 0xF3, 0xB7, 0x6B, 0x22, 0x2B, 0xBD, 0x69, 0x42, 0x4B, 0xA5, 0xEA, 0xA6, 0xD2, 0x6F, 0x4F,
    0x4E, 0x07, 0xE1, 0x36, 0x01, 0xB5, 0xAA, 0xB1, 0x94, 0x0B, 0x35, 0x3A, 0xC7, 0x49, 0x53, 0x82,
    0xC3, 0x7B, 0x32, 0xFF, 0x19, 0xC4, 0xF1, 0xC9, 0xE8, 0xF7, 0x56, 0x15, 0xA3, 0x46, 0x89, 0x43,
    0x9D, 0x8F, 0x20, 0xEF, 0xBB, 0x2A, 0xCB, 0x09, 0x93, 0x4A, 0x1C, 0xE3, 0x33, 0xD1, 0xE0, 0x1D,
    0x72, 0x7C, 0x27, 0xE9, 0x17, 0x28, 0x6D, 0x6A, 0xD9, 0x00, 0x9A, 0xE5, 0x63, 0xDE, 0x23, 0x9F,
    0x0D, 0x47, 0x3B, 0x65, 0x08, 0x84, 0x6C, 0x1A, 0x88, 0x12, 0xA1, 0xA4, 0xB3, 0x18, 0x24, 0x1B,
    0xD7, 0x44, 0xDB, 0xAC, 0x6E, 0x7D, 0x51, 0x5E, 0xED, 0x50, 0xD6, 0x11, 0x5B, 0x9C, 0xB4, 0x68,
    0x3D, 0x2F, 0x03, 0x40, 0xBA, 0x2E, 0xCA, 0x02, 0xE6, 0xA8, 0xEC, 0x83, 0x06, 0x5D, 0xB8, 0x4D,
    0x97, 0x66, 0xF0, 0xFB, 0x8A, 0x55, 0xAB, 0xB2, 0x04, 0xFA, 0x0A, 0x31, 0x71, 0xCC, 0x8B, 0x73,
    0xA9, 0x48, 0x5C, 0xF9, 0x98, 0xE4, 0xC6, 0x34, 0xC5, 0x7E, 0x81, 0x75, 0x90, 0x1F, 0x92, 0x3F,
    0x9E, 0x10, 0x29, 0x52, 0x39, 0xF4, 0x41, 0x78, 0x5F, 0x16, 0x79, 0xC2, 0xB0, 0xDD, 0xF2, 0x61,
    0x0F, 0x70, 0xD4, 0x91, 0xDC, 0xF6, 0xF8, 0xFD, 0x59, 0x38, 0x8D, 0x96, 0xAE, 0x8E, 0x76, 0xA2,
}
 
// 枚举 256 就能得出源 byte
func dec(cmp byte, idx int) byte {
    // 第三部分算法
    enc := func(x byte) byte {
        c := t2tb2[idx]
        for i := 0; i < 8; i++ {
            if x&1 != 0 {
                c = c + 1
            } else {
                c = t2tb2[c]
            }
            x = x >> 1
        }
        if idx == 0 || idx == 7 {
            c--
        }
        return c
    }
    for i := 0; i < 256; i++ {
        if enc(byte(i)) == cmp {
            return byte(i)
        }
    }
    return 0
}
 
// 求解第二部分
func t2p2() {
    var arr []byte
    for i, c := range []byte("GoodJob~") {
        arr = append(arr, dec(c, i))
    }
    printHex(arr)
}
 
func main() {
    t2p1()
    t2p2()
}
package main
 
import (
    "fmt"
)
 
// tb 404000
var t2tb = []byte{
    0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0xA2, 0x9B,
    0xF4, 0xDF, 0xAC, 0x7C, 0xA1, 0xC6, 0x16, 0xD0, 0x0F, 0xDD,
    0xDC, 0x73, 0xC5, 0x6B, 0xD1, 0x96, 0x47, 0xC2, 0x26, 0x67,
    0x4E, 0x41, 0x82, 0x20, 0x56, 0x9A, 0x6E, 0x33, 0x92, 0x88,
    0x29, 0xB5, 0xB4, 0x71, 0xA9, 0xCE, 0xC3, 0x34, 0x50, 0x59,
    0xBF, 0x2D, 0x57, 0x22, 0xA6, 0x30, 0x04, 0xB2, 0xCD, 0x36,
    0xD5, 0x68, 0x4D, 0x5B, 0x45, 0x9E, 0x85, 0xCF, 0x9D, 0xCC,
    0x61, 0x78, 0x32, 0x76, 0x31, 0xE3, 0x80, 0xAD, 0x39, 0x4F,
    0xFA, 0x72, 0x83, 0x4C, 0x86, 0x60, 0xB7, 0xD7, 0x63, 0x0C,
    0x44, 0x35, 0xB3, 0x7B, 0x19, 0xD4, 0x69, 0x08, 0x0B, 0x1F,
    0x3D, 0x11, 0x79, 0xD3, 0xEE, 0x93, 0x42, 0xDE, 0x23, 0x3B,
    0x5D, 0x8D, 0xA5, 0x77, 0x5F, 0x58, 0xDB, 0x97, 0xF6, 0x7A,
    0x18, 0x52, 0x15, 0x74, 0x25, 0x62, 0x2C, 0x05, 0xE8, 0x0D,
    0x98, 0x2A, 0x43, 0xE2, 0xEF, 0x48, 0x87, 0x49, 0x1C, 0xCA,
    0x2B, 0xA7, 0x8A, 0x09, 0x81, 0xE7, 0x53, 0xAA, 0xFF, 0x6F,
    0x8E, 0x91, 0xF1, 0xF0, 0xA4, 0x46, 0x3A, 0x7D, 0x54, 0xEB,
    0x2F, 0xC1, 0xC0, 0x0E, 0xBD, 0xE1, 0x6C, 0x64, 0xBE, 0xE4,
    0x02, 0x3C, 0x5A, 0xA8, 0x9F, 0x37, 0xAF, 0xA0, 0x13, 0xED,
    0x1B, 0xEC, 0x8B, 0x3E, 0x7E, 0x27, 0x99, 0x75, 0xAB, 0xFE,
    0xD9, 0x3F, 0xF3, 0xEA, 0x70, 0xF7, 0x95, 0xBA, 0x1D, 0x40,
    0xB0, 0xF9, 0xE5, 0xF8, 0x06, 0xBC, 0xB6, 0x03, 0xC9, 0x10,
    0x9C, 0x2E, 0x89, 0x5C, 0x7F, 0xB1, 0x1A, 0xD6, 0x90, 0xAE,
    0xDA, 0xE6, 0x5E, 0xB9, 0x84, 0xE9, 0x55, 0xBB, 0xC7, 0x0A,
    0xE0, 0x66, 0xF2, 0xD8, 0xCB, 0x00, 0x12, 0xB8, 0x17, 0x94,
    0x6A, 0x4A, 0x01, 0x24, 0x14, 0x51, 0x07, 0x65, 0x21, 0xC8,
    0x38, 0xFD, 0x8F, 0xC4, 0xF5, 0xFC,
}
 
func getloss() (ret []byte) {
    m := make(map[byte]bool)
    for i := 8; i < 256; i++ {
        m[t2tb[i]] = true
    }
 
    for i := 0; i < 256; i++ {
        if !m[byte(i)] {
            ret = append(ret, byte(i))
        }
    }
    return
}
 
// 排列算法
func permutations(arr []byte) [][]byte {
    var helper func([]byte, int)
    res := [][]byte{}
 
    helper = func(arr []byte, n int) {
        if n == 1 {
            tmp := make([]byte, len(arr))
            copy(tmp, arr)
            res = append(res, tmp)
        } else {
            for i := 0; i < n; i++ {
                helper(arr, n-1)
                if n%2 == 1 {
                    tmp := arr[i]
                    arr[i] = arr[n-1]
                    arr[n-1] = tmp
                } else {
                    tmp := arr[0]
                    arr[0] = arr[n-1]
                    arr[n-1] = tmp
                }
            }
        }
    }
    helper(arr, len(arr))
    return res
}
 
// 第一个校验算法
func checksk(c [256]byte) bool {
    var sss [256 * 2]byte
    var kkk [256][256]byte
    maxn := []int{2, 4, 8, 0x10, 0x20, 0x40, 0x80}
    for i := 0; i < 256; i++ {
        sss[0] = c[i]
        sss[1] = byte(i + 1)
        var ps int
        k := maxn[0]
        for _, n := range maxn {
            for j := 0; j < n; j++ {
                sss[k] = c[sss[ps]]
                sss[k+1] = sss[ps] + 1
                k += 2
                ps++
            }
        }
 
        for j := 0; j < 256; j++ {
            kkk[i][sss[ps]]++
            ps++
        }
    }
 
    var r1, r2, r3, r4 int
    for i := 0; i < 256; i++ {
        if kkk[i][0] != 0 {
            r1++

[招生]科锐逆向工程师培训(2024年11月15日实地，远程教学同时开班, 第51期)

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