for i<-1 to n do
for j<-1 to n^2 do
write(i*j)
end-for
end-for
corby
-4
Junior Poster
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Jump to Postcall clock() before the loop starts and again after the loop ends, then subtract the two times. When the program runs fast enough the difference might be 0. In that case, put the loops into a function and call that function 1,000 times.
clock_t t1, t2; t1 …
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