Answers for "Write a program to accept 10 integers in an array and for each one of them, separately, indicate the number of even digits, odd digits, and total number of digits"

Print Odd Even Negative Integer Count

#include<stdio.h>
int main()
{
int n,i,e=0,o=0,ne=0;
scanf("%d",&n);
int a[n];
for(i=0;i<n;i++){
    scanf("%d",&a[i]);
if(a[i]<0) ne++;
else if(a[i]%2==0&&a[i]>=0) e++;
else o++;
}
    printf("%d %d %d",o,e,ne);

}

Posted by: Guest on July-09-2020

Source

Print Odd Even Negative Integer Count

n,neg,odd,eve,m=int(input()),0,0,0,[int(i) for i in input().split()]
for i in m:
    if i<0:
        neg+=1
    elif i%2==0 :
        eve+=1
    else:
        odd+=1
print("{0} {1} {2}".format(odd,eve,neg))

Posted by: Guest on July-09-2020

Source

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Answers for "Write a program to accept 10 integers in an array and for each one of them, separately, indicate the number of even digits, odd digits, and total number of digits"

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