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#include using namespace std; void print(int marks_arr[],int cnt) { int ind[cnt]; int i=0; int j=0; int k=0; int s=0; for(k=0;k {ind[k]=0; } int cnt1=0; for(i=0;i { ind[cnt1]=1; for(j=i+1;j { //cout< if((marks_arr[i]==marks_arr[j])&&(marks_arr[i]!=-1)) { ind[cnt1]=ind[cnt1]+1; marks_arr[j]=-1; //if(marks_arr[j]==-1) //{ //break; //}
} } cnt1++; } for(s=0;s { if(marks_arr[s]!=-1) { cout<<"\n"< }
}
} int add() { int mark; cout<<"Enter the mark:"; cin>> mark; return mark; } float avg(int mark[],int c) { int i,sum=0; float average; for(i=0;i { sum= sum+mark[i]; } average=sum/c; return average; } int marks_arr[100]; int main() { static int count=0; //int marks_arr[100], int choice,i,ch,searchelement,totalcount,total; float val; do { cout<<"\n 1. Add student marks"; cout<<"\n 2. Display student marks"; cout<<"\n 3. Find average of all marks"; cout<<"\n 4. Calculate the standard deviation"; cout<<"\n 5. Delete a student mark"; cout<<"\n 6. Search all marks "; cout<<"\n 7. Print all marks";cout<<"\nEnter choice : "; cin>>choice; switch(choice) { case 1: { marks_arr[count]=add(); count++; break; } case 2: { for(i=0;i { cout< } break; } case 3: { val= avg(marks_arr,count); cout<< "The avg of total marks is: "< break; } case 4: case 5: case 6: { cout<<"\n Enter the mark to search:"; cin>>searchelement; totalcount=0; for(i=0;i { if(marks_arr[i]==searchelement) { totalcount=totalcount+1; } } cout<<"No of times the element is present : "< break; }case 7: { print(marks_arr,count); } } cout<<"\nContinue [0-yes / 1-no] :"; cin>>ch; }while(ch!=1); }
C Program for DIVISER void main() { int result,number,min; clrscr(); printf("ENTER THE NUMBER="); flushall();
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Using Figure 10.2 as a model, illustrate the result of each operation in the sequence ENQUEUE.Q; 4/, ENQUEUE.Q; 1/, ENQUEUE.Q; 3/, DEQUEUE.Q/, ENQUEUE.Q; 8/, and DEQUEUE.Q/ on an i
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A company needs 200 pencils per year . you cannot simply use this price as the cost of pencils two years from now. Because of inflation the cost is likely to be higher than it is
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