Consider the same code, copied here for reference, but clearly with an array that has no valid sequence: int Length 10; = int Seq[] = {-10, 0, 1, 500, 7, 11, -34, 14, 1024, 18}; int Diff; int i, j; int main() { for (i=0; i

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Please answer the question in C code

Consider the same code, copied here for reference, but clearly with an array that has no valid
sequence:
int Length
=
10;
int Seq[] = {-10, 0, 1, 500, 7, 11, -34, 14, 1024, 18};
int Diff;
int i, j;
int main() {
for (i=0; i<Length-2; i++) {
Diff =
Seq[i+1]-Seq[i]; // The difference, if it's an arithmetic sequence
for (j=i+1; j<Length-1; j++) {
if ( (Seq[j+1]-Seq[j])!= Diff) break; // Quit trying, if an element fails
}
if (j==9) {
printf("The arithmetic sequence starts at Seq[%d]\n", i);
break;
}
}
if (i =8) {
printf("There is no arithmetic sequence.\n");
}
return 0;
}
In the box below, type two numbers, representing what should be in the two blanks below (in the
same order).
The statement
Diff Seq[i+1]-Seq[i];
will be executed
times, and the statement
if ( (Seq[j+1]-Seq[j]) != Diff)
will be executed
times
Transcribed Image Text:Consider the same code, copied here for reference, but clearly with an array that has no valid sequence: int Length = 10; int Seq[] = {-10, 0, 1, 500, 7, 11, -34, 14, 1024, 18}; int Diff; int i, j; int main() { for (i=0; i<Length-2; i++) { Diff = Seq[i+1]-Seq[i]; // The difference, if it's an arithmetic sequence for (j=i+1; j<Length-1; j++) { if ( (Seq[j+1]-Seq[j])!= Diff) break; // Quit trying, if an element fails } if (j==9) { printf("The arithmetic sequence starts at Seq[%d]\n", i); break; } } if (i =8) { printf("There is no arithmetic sequence.\n"); } return 0; } In the box below, type two numbers, representing what should be in the two blanks below (in the same order). The statement Diff Seq[i+1]-Seq[i]; will be executed times, and the statement if ( (Seq[j+1]-Seq[j]) != Diff) will be executed times
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