If $C$ is an integer, no cuts are required.
Otherwise let $c=C-\lfloor C\rfloor$, a real number between $0$ and $1$.
Each rope must finally contain at least one cut end, thus the number of cuts is at least $\frac N2$ and it is easily solvable with $N$ cuts.
Indeed, $\lceil \frac N2 \rceil$ is enough if $c=\frac12$.
If $c=\frac23$, one can produce $\lceil \frac23 N\rceil$ pieces of $\frac23$ and combine the $\frac13$ rests for the remaining ropes, hence $\lceil \frac23 N\rceil$ cuts suffice.
If $c=\frac13$, then $\lceil \frac23 N\rceil$ cuts suffice again.
For $c=\frac34$ or $c=\frac14$, I can do it with $\lceil \frac34 N\rceil$ cuts, for $\frac k 5$ with $1\le k\le 4$, I can do it with $\lceil \frac45N\rceil$ cuts.
A pattern sems to emerge here, but I'm not sure if it is really optimal: $c=\frac pq$ requires $\lceil (1-\frac1q)N\rceil$ cuts. And if $c$is irrational, $N$ cuts are needed, I suppose.
Remark: If the final result is correct, there is no need to go for $c$, one can directly say that $\lceil (1-\frac1q)N\rceil$ cuts are needed if $\frac CR=\frac pq$.