Answer:
5,10; 6,12; 7,14
Explanation:
We will demonstrate the iteration of the loop:
First iteration: Number = 7, Count = 5 at the beginning. We will check if Count <= Number? Since it is correct, prints 5,10. Increment the Count by 1.
Second iteration: Number = 7, Count = 6. We will check if Count <= Number? Since it is correct, prints 6,12. Increment the Count by 1.
Third iteration: Number = 7, Count = 7. We will check if Count <= Number? Since it is correct, prints 7,14. Increment the Count by 1.
Forth iteration: Number = 7, Count = 8. We will check if Count <= Number? Since it is not correct, the loop stops.
Answer:
#include <stdio.h>
#include <ctype.h>
void printHistogram(int counters[]) {
int largest = 0;
int row,i;
for (i = 0; i < 26; i++) {
if (counters[i] > largest) {
largest = counters[i];
}
}
for (row = largest; row > 0; row--) {
for (i = 0; i < 26; i++) {
if (counters[i] >= row) {
putchar(254);
}
else {
putchar(32);
}
putchar(32);
}
putchar('\n');
}
for (i = 0; i < 26; i++) {
putchar('a' + i);
putchar(32);
}
}
int main() {
int counters[26] = { 0 };
int i;
char c;
FILE* f;
fopen_s(&f, "story.txt", "r");
while (!feof(f)) {
c = tolower(fgetc(f));
if (c >= 'a' && c <= 'z') {
counters[c-'a']++;
}
}
for (i = 0; i < 26; i++) {
printf("%c was used %d times.\n", 'a'+i, counters[i]);
}
printf("\nHere is a histogram:\n");
printHistogram(counters);
}
Answer:
The term "savanna" is often used to refer to open grassland with some tree cover, while "grassland" refers to a grassy ecosystem with little or no tree cover.
Explanation:
Answer:
I am guessing animal crossing
Answer:
Derive FROM invoice_transaction, invoice_details, item_details
and JOIN customer_details ON (invoice_transaction.CUST_ID = customer_details.CUST_ID AND customer_details.FIRST_NAME = 'James' AND customer_details.LAST_NAME = 'Gonzalez')
Explanation:
The following details will be there in the invoice
-
item_details
- rep_details
- invoice_details
- customer_details
- invoice_transaction
Derive FROM invoice_transaction, invoice_details, item_details
and JOIN customer_details ON (invoice_transaction.CUST_ID = customer_details.CUST_ID AND customer_details.FIRST_NAME = 'James' AND customer_details.LAST_NAME = 'Gonzalez')
