1.calculates and prints the number of and sum of the integers between 1 and 100 (inclusive) that are not divisible by 3 or 5
codes:
#include
int main(){
int i;
int sum=0;
int count=0;
for(i=1;i<=100;i++){
if(i%3!=0 && i%5!=0){
sum=sum+i;
count=count+1;
}
}
printf("sum=%d number=%d\n",sum,count);
return 0;
}
2.The program prompts for a string (max 80 characters) and converts uppercase characters to lowercase
codes:
#include
#include
int main(void) {
char str[81];
int i, len;
printf("Enter string:\n");
fgets(str, 81, stdin);
len = strlen(str);
if(str[len-1] == '\n') {
str[len-1] = 0;
}
for(i=0; str[i]; i++) {
if(str[i] >= 'A' && str[i] <= 'Z') {
str[i] += 32;
}
}
printf("%s\n", str);
return 0;
}
3.Convert lowercase characters to uppercase instead.
codes:
#include
#include
int main(void) {
char str[81];
int i, len;
printf("Enter string:\n");
fgets(str, 81, stdin);
len = strlen(str);
if(str[len-1] == '\n') {
str[len-1] = 0;
}
for(i=0; str[i]; i++) {
if(str[i] >= 'a' && str[i] <= 'z') {
str[i] -= 32;
}
}
printf("%s\n", str);
return 0;
}
4.Count the number of each type of letter (A to Z) and print a table showing these counts. Ignore non-alphabetic characters.
codes:
#include
#include
int main(void) {
char str[81];
int lettercount[26];
int i, len;
for(i=0; i<26; i++) {
lettercount[i] = 0;
}
printf("Enter string:\n");
fgets(str, 81, stdin);
len = strlen(str);
if(str[len-1] == '\n') {
str[len-1] = 0;
}
for(i=0; str[i]; i++) {
if(str[i] >= 'A' && str[i] <= 'Z') {
lettercount[str[i]-'A']++;
} else if(str[i] >= 'a' && str[i] <= 'z') {
lettercount[str[i]-'a']++;
}
}
for(i=0; i<26; i++) {
printf("%c: %d\n", 'A'+i, lettercount[i]);
}
return 0;
}
5.Accept 20 lines of text (each up to 80 characters) instead of 1 line, then print a count of each letter seen (the total over all lines).
codes:
#include
#include
int main(void) {
char str[81];
int lettercount[26];
int i, len;
int linecount;
for(i=0; i<26; i++) {
lettercount[i] = 0;
}
for(linecount=0; linecount<20; linecount++) {
printf("Enter string:\n");
fgets(str, 81, stdin);
len = strlen(str);
if(str[len-1] == '\n') {
str[len-1] = 0; /* Null character - we could also have said '\0' */
}
for(i=0; str[i]; i++) {
if(str[i] >= 'A' && str[i] <= 'Z') {
lettercount[str[i]-'A']++;
} else if(str[i] >= 'a' && str[i] <= 'z') {
lettercount[str[i]-'a']++;
}
}
}
for(i=0; i<26; i++) {
printf("%c: %d\n", 'A'+i, lettercount[i]);
}
return 0;
}
6.As 5, but the program accepts any number of lines of text, up to when it sees a line containing only the text “END” (without the quotes)
codes:
#include
#include
int main(void) {
char str[81];
int lettercount[26];
int i, len;
int finished=0;
for(i=0; i<26; i++) {
lettercount[i] = 0;
}
while(!finished) {
printf("Enter string:\n");
fgets(str, 81, stdin);
len = strlen(str);
if(str[len-1] == '\n') {
str[len-1] = 0;
}
if(strcmp(str, "END") == 0) {
break;
}
for(i=0; str[i]; i++) {
if(str[i] >= 'A' && str[i] <= 'Z') {
lettercount[str[i]-'A']++;
} else if(str[i] >= 'a' && str[i] <= 'z') {
lettercount[str[i]-'a']++;
}
}
}
for(i=0; i<26; i++) {
printf("%c: %d\n", 'A'+i, lettercount[i]);
}
return 0;
}