Coin Ecryption SOLUTIONS BINCOIN
Culinary expert and Ada have been alloted an errand of finding a mystery paired code which has been encoded in a strange way. There are N and M coins covered up in the ways followed by Chef and Ada individually. Each coin has a number composed on one side while opposite side contains a letters in order. Gourmet expert and Ada begin finding the coins from the beginning stages of their particular ways. In this cycle they probably won’t have the option to discover a few coins and push forward.
There is an uncommon possibility that the all the coins picked and the request for picking them will be comparative for Chef and Ada regarding numbers composed on them. Let the most number of coins that these two can get and fulfill this uncommon condition be K. The mystery double code is the twofold portrayal of XOR of ASCII portrayals of the letters in order composed on coins at area K on the ways of Chef and Ada separately.
The main line of the info contains a solitary number T meaning the quantity of experiments. The portrayal of T experiments follows.
The primary line of each experiment contains two space-isolated whole numbers N and M.
The second line of each experiment contains N space isolated numbers C1,C2,C3,… CN.
The third line of each experiment contains a line of length N.
The fourth line of each experiment contains M space isolated numbers A1,A2,A3,… AN.
The fifth line of each experiment contains a line of length M.
Note: The letters in order at position I in the string compares to the whole number at position I in the succession.
For each expeiment print a parallel string – double portrayal of XOR of ASCII estimations of letter sets at area K.
Note : Output of each experiment ought to be imprinted on new line.
1 2 3
2 3 4 5
In the best case Chef and Ada can share coins [2,3] for all intents and purpose for example 2 comparable coins subsequently the parallel code is the XOR of ASCII estimation of second component of both the strings for example ASCII estimation of b = 98, q = 113 consequently we will find our last solution by doing 98 XOR 113.
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