angstromctf-2019-部分WriteUp

<canvas id="canvas" width="256" height="256">对不你，你的浏览器不支持Canvas</canvas>
<script>

var ctxt = canvas.getContext('2d');
var img = new Image(256,256);
img.src = 'img.png';//src也可以是从文件选择控件中取得。
img.onload = function(){
    ctxt.drawImage(img, 0, 0);
    var data = ctxt.getImageData(0, 0, 256, 256).data;//读取整张图片的像素。
	var pixel = [0,0,0]//像素RGB
	var flag = new Array(256)
	for (var i=0;i<256;i++)
	{

		var chr = 0;//用来存储解密出来的字符
		//k是第k位     读出来的图片自动加了A通道
		for (var j=0,k=0;j<256;j=j+32,k++)
		{
			pixel[0] = data[((i * (256 * 4)) + (j*4)) + 0]; //R值
			pixel[1] = data[((i * (256 * 4)) + (j*4)) + 1]; //G值
			pixel[2] = data[((i * (256 * 4)) + (j*4)) + 2]; //B值
			RxorG = pixel[0]^pixel[1];
			if(pixel[2]%2==0){
				if(RxorG&1==1){
					chr = chr | (1 << k)
				}else{
					chr = chr | (0 << k)
				}
			}else{
				if(RxorG&1==1){
					chr = chr | (0 << k)
				}else{
					chr = chr | (1 << k)
				}
			}
			
		}
		flag[i] = chr;
	}
	console.log(flag);
	var ascii_flag = '';
	for (var m=0;m<flag.length;m++){
		ascii_flag = ascii_flag + String.fromCharCode(flag[m]);
	}
	console.log(ascii_flag);
	
}

</script>

from PIL import Image
image = Image.open('out.png')
pixels = list(image.getdata())
flag = [0 for _ in range(256)]
for i in range(256):
	row = pixels[i*256:(i+1)*256]
	for j in range(8):
		pixel = row[j*32]
		bits = [x & 1 for x in pixel]
		bit = bits[0] ^ bits[1] ^ bits[2]
		flag[i] ^= bit << j
flag = bytes(flag)
print(flag)

def decrypt(d, s):
	e = ''
	for c in d:
		e += chr((ord(c)+s) % 0xff)
	return e
enc = ':<M?TLH8<A:KFBG@V'
for i in range(256):
	print decrypt(enc,i)

p = 8337989838551614633430029371803892077156162494012474856684174381868510024755832450406936717727195184311114937042673575494843631977970586746618123352329889
q = 7755060911995462151580541927524289685569492828780752345560845093073545403776129013139174889414744570087561926915046519199304042166351530778365529171009493
e = 65537
c = 7022848098469230958320047471938217952907600532361296142412318653611729265921488278588086423574875352145477376594391159805651080223698576708934993951618464460109422377329972737876060167903857613763294932326619266281725900497427458047861973153012506595691389361443123047595975834017549312356282859235890330349

import gmpy2
import binascii
p = gmpy2.mpz(8337989838551614633430029371803892077156162494012474856684174381868510024755832450406936717727195184311114937042673575494843631977970586746618123352329889)
q = gmpy2.mpz(7755060911995462151580541927524289685569492828780752345560845093073545403776129013139174889414744570087561926915046519199304042166351530778365529171009493)
e = gmpy2.mpz(65537)
phi_n = (p - 1) * (q - 1)
d = gmpy2.invert(e, phi_n)
print ("private key:")
print (d)

c = gmpy2.mpz(7022848098469230958320047471938217952907600532361296142412318653611729265921488278588086423574875352145477376594391159805651080223698576708934993951618464460109422377329972737876060167903857613763294932326619266281725900497427458047861973153012506595691389361443123047595975834017549312356282859235890330349)
print ("plaintext:")
m = pow(c,d,p*q)
print m
m_hex = hex(m)[2:]
print("ascii:\n%s"%(binascii.a2b_hex(m_hex).decode("utf8"),))

from secret import flag

def xor(x, y):
	o = ''
	for i in range(len(x)):
		o += chr(ord(x[i])^ord(y[i]))
	return o

assert len(flag) % 2 == 0

half = len(flag)//2
milk = flag[:half]
cream = flag[half:]

assert xor(milk, cream) == '\x15\x02\x07\x12\x1e\x100\x01\t\n\x01"'

n: 99157116611790833573985267443453374677300242114595736901854871276546481648883
g: 99157116611790833573985267443453374677300242114595736901854871276546481648884
c: 2433283484328067719826123652791700922735828879195114568755579061061723786565164234075183183699826399799223318790711772573290060335232568738641793425546869