You and three of your friends decide to hike the Appalachian Trail, a 2175-mile trail running from Georgia to Maine, starting wh
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Answer:
Total number of character comparison = 43
Explanation:
Using the Brute force algorithm
The string of n characters is known as text, and the string of m characters is known as the pattern.
From the given information:
The text (n)=THERE_IS_MORE_TO_LIFE_THAN_INCREASING_ITS_SPEED
The pattern (m) = GANDHI
The total no of characters that we have in the text = 47
The total number of characters in pattern = 6
For a brute force algorithm;
Since; the first character of the pattern does not exist in the text, then the number of trials made can be attempted can be expressed as = n – m + 1
= 47 – 6 + 1
= 47 – 5
= 42
Thus; the algorithm will attempt the trial 42 times.
Now, for loop in the algorithm to run 42 times, the G in the pattern will have to align against the for T in the text, and in the last case, it will be aligned against the last space.
On each attempted trial, the algorithm will make one unsuccessful comparison.
However, at the trial at which the G in the pattern Is aligned with the G in the text, there will be two successful comparisons.
Hence, we can calculate the total number of character comparison as follows:
Total number of character comparison =
Total number of character comparison = ( (( 42 – 1) × 1 ) + ( 1 × 2) )
Total number of character comparison = 41 + 2
Total number of character comparison = 43
Answer:
See the code below and the algorithm explanation on the figure.
Explanation:
The explanation in order to get the answer is given on the figure below.
Solving this problem with C. The program is given below:
#include <stdio.h>
int main(void) {
int n, Even=0, Odd=0, Zeros=0;
for (;;) {
printf("\nEnter the value the value that you want to check(remember just integers): ");
//IF we input a non-numeric character the code end;
if (scanf("%d", &n) != 1) break;
if (n == 0) {
Zeros++;
}
else {
if (n % 2) {
Even++;
}
else {
Odd++;
}
}
}
printf("for this case we have %d even, %d odd, and %d zero values.", Even, Odd, Zeros);
return 0;
}
