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22 lines
580 B
22 lines
580 B
from Crypto.Util.number import inverse
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from fractions import *
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e = 43
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n = 2531257
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c = "906851 991083 1780304 2380434 438490 356019 921472 822283 817856 556932 2102538 2501908 2211404 991083 1562919 38268"
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c = c.split(" ")
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c = map(int, c)
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p = 509
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q = 4973
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phi = (p - 1) * (q - 1)
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d = inverse(e, phi)
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m = []
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for i in c:
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m.append(pow(i, d, n))
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print m
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plain = [103, 105, 103, 101, 109, 123, 83, 97, 118, 97, 103, 101, 95, 83, 105, 120, 95, 70, 108, 121, 105, 110, 103, 95, 84, 105, 103, 101, 114, 115, 125]
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flag = ""
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for i in plain:
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flag += chr(i)
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print "Decoded flag : ", flag
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