Quote:
I'm not sure if this can actually be done.
It can be done.
Many approach are possible.
My approach would be to scale the numbers until they are all integers, then use big integers. Because integer arithmetic is faster than floating point and big integer are very common as of today.
0.199848000000000 X 4.974380000000000 = 0.99412293432154732337566954206
1.99848000000000 X 49.74380000000000 = 99.412293432154732337566954206
19.9848000000000 X 497.4380000000000 = 9941.2293432154732337566954206
199.848000000000 X 4974.380000000000 = 994122.93432154732337566954206
1998.48000000000 X 49743.80000000000 = 99412293.432154732337566954206
199848.000000000 X 497438.0000000000 = 9941229343.2154732337566954206
...
1998480000000000 X 4974380000000000 = 994122934321547323375669542060
then solve P and Q in
(1998480000000000 + P) X (4974380000000000 + Q) = 994122934321547323375669542060
I would try to factorize 994122934321547323375669542060
994122934321547323375669542060 = 2^2 × 3^4 × 5 × 7^3 × 103 × 63577 × 69203 × 3947923974637
Quote:
I have to admit my ignorance and confess I do not know anything about programming, yet I really need this one solved for me.
You have been given general guidelines because it is what can be done in the scope of this forum.
Writing efficient code for this problem is a real job because one need to do a real study to avoid pitfalls, and it takes time.