ex37

 

Write a program that allows the entrance of 10 integers in a table. Then count how many Odd and Even elements are there. 

******** In Algorithm ********

Algorithm Variables i, tab[10],Nb_pair, Nb_odd :entiers Begin Nb_pair ← 0 Nb_odd ← 0 for i from 1 to 10 write("Enter an integer :") read(tab[i]) next for i from 1 to 10 if( tab[i] % 2 = 0 ) then Nb_even ← Nb_even + 1 else Nb_odd ← Nb_odd + 1 next next write("Number of even = ",Nb_even) write("Number of odd = ",Nb_odd) End Result ==> 1 4 3 6 6 9 0 9 7 1 Number of even = 4 Number of odd = 6

******** In C ********

#include<stdio.h> int main(){ int i,tab[10],Nb_even=0,Nb_odd=0; for(i=0;i<10;i++){ printf("Enter an integer : "); scanf("%d",&tab[i]); } for(i=0;i<10;i++){ if( tab[i]%2==0 ) Nb_even ++; else Nb_odd++ ; } printf("Number of even = %d\n",Nb_even); printf("Number of odd = %d",Nb_odd); return 0; }

******** In C++ ********

#include <iostream> using namespace std; int main(){ int i,tab[10],Nb_even=0,Nb_odd=0; for(i=0;i<10;i++){ cout<<"Enter an integer :"; cin>>tab[i]; } for(i=0;i<10;i++){ if( tab[i]%2==0 ) Nb_even ++; else Nb_odd++ ; } cout<<"Number of even = "<<Nb_even<<endl; cout<<"Number of odd ="<<Nb_odd; return 0; }