RDO.
RDO uses the lower-level DAO and ODBC for direct access to databases.
(Disclaimer: I am not a professional, so it might not be the most concise answer possible, but I did run the Python script and it works)
Answer:
user_input = input("What food do you have in your refrigerator? ").lower()
count = 0
while True:
if user_input != 'apples':
count += 1
print(f'You have a {user_input} with a total of {count} food(s)\n')
user_input = input("What food do you have in your refrigerator? ")
else:
break
Answer: It allows us to do everyday tasks on the internet
Explanation: We wouldn’t be able to email, research, etc without coding!
Answer:
hope this helps and do consider giving a brainliest to the ans if it helped.
Explanation:
//program to check if the entered grid is magic square or not
/**c++ standard libraries
*/
#include<bits/stdc++.h>
using namespace std;
/**function to check whether the entered grid is magic square or not
*/
int isMagicSquare(int arr[3][3]){
int i,j,sum=0,sum1=0,rsum,csum;
for(i=0;i<3;i++){
sum+=arr[i][i];
sum1+=arr[i][2-i];
}
if(sum!=sum1){
return 0;
}
for(i=0;i<3;i++){
rsum=0;
csum=0;
for(j=0;j<3;j++){
rsum+=arr[i][j];
csum+=arr[j][i];
}
if(sum!=rsum){
return 0;
}
if(sum!=csum){
return 0;
}
}
return 1;
}
/** main function to get user entries and
* call function
* and print output
*/
int main(){
int i,j,arr[3][3]={0};
for(i=0;i<3;i++){
for(j=0;j<3;j++){
cout<<"Enter the number for row "<<i<<" and column "<<j<<" : ";
cin>>arr[i][j];
}
}
int ret = isMagicSquare(arr);
if(ret==1){
cout<<"This is a Lo Shu magic square"<<endl;
}
else{
cout<<"This is not a Lo Shu magic square"<<endl;
}
return 0;
}
