All generations.
def line_multiples(n):
for i in range(1, n+1):
for j in range(1, n+1):
if i*j == n:
yield i, j
def line_multiples_prime(n):
for i in range(1, n+1):
for j in range(1, n
id printLine(int n){
for(int i = 0; i < n; i++){
cout << "-";
}
cout << endl;
}
int main(){
int n = 0;
cin >> n;
printLine(n);
return 0;
}
A:
You can use a for loop to print the lines. The loop will run n
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