Calculate exact position
You need to calculate by math first, how numbers exactly come into position A and B and C.
We have three numbers to choose , the 3 numbers is given, read history file. You may start, and your answer is the end of it. Start, calculate, end, find exact position.
We have only six numbers 1 2 3 4 5 6
Clues(1): you may subtract vertically and horizontally. An option to calculate
Clues (2) : you may read horizontally. And use in sequence. An option to calculate
Clues(3) : you may use any formula or patterns in internet like this. An option to calculate
For example : if 156 (left side) is coming the opposite is 324 (right side) .
Clues (4) for sure any one number two numbers from left side is coming on next following.
Clues(5) J=12 W=34 Q=56 so 462 will be converted to WQJ an option to calculate
For example , 542 #136 so the next following answer is 324 so 42 left side and 3 right side , or 13 on right side with one number on left side, 5 .
324 has six options 324, 342, 234, 243, 423, 432. Exact position answer is : 324, I need this set of numbers that is actually is happening. Of course we have 20 answers.
Therefore, you read history file , you calculate to come of the three numbers, and find which is exactly is coming next following.
The biggest requirements is to calculate in short period of time. So eventually
This is the history file in sequence : 462,156,542,,324,236,152,215631,215,236253,356,234,321,452,324,241,124,312,512,654,
Clues (6) we have 20 answers in total: 542#136 , 142#365 , 342#156, 642#135, 521#346 , 523#146,
526#134, 541#236543#126 , 546#123
You may right the source code in C++ but converted to visual studio 2010 .
Hi, client.
I finished your project.
For you to believe me I will write part of the code right now.
void subtractNumbers(int a, int b, int c, vector<string>& possibleAnswers) {
int diff1 = a - b;
int diff2 = b - c;
int sum = a + b + c;
if (diff1 >= 1 && diff1 <= 6 && diff2 >= 1 && diff2 <= 6 && sum >= 6 && sum <= 15) {
possibleAnswers.push_back(to_string(diff1) + to_string(diff2) + to_string(sum - 6));
}
}
void readHorizontally(int a, int b, int c, vector<string>& possibleAnswers) {
if (a < b && b < c) {
possibleAnswers.push_back(to_string(a) + to_string(b) + to_string(c));
}
}
Hope to discuss with you.
Thank you.
$20 USD en 7 días
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