Embedded Files
1. Read numbers from stdin and print their sum to stdout.
Sol:
#include <iostream>
using namespace std;
int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int a, b, c;
cin >> a >> b >> c;
cout << a + b + c;
return 0;
}
2. Input consists of the following space-separated values: int, long, char, float, and double, respectively.
Sol:
int main() {
int i;
long int li;
char c;
float f;
double d;
cin >> i >> li >> c >> f >> d;
cout << i << endl << li << endl << c << endl;
printf("%0.3f\n",f);
printf("%0.9lf",d);
// Complete the code.
return 0;
}
3. Given a positive integer , do the following: If , print the lowercase English word corresponding to the number (e.g., one for , two for , etc.).If , print Greater than 9..
Sol:
int main()
{
string n_temp;
getline(cin, n_temp);
int n = stoi(ltrim(rtrim(n_temp)));
if ( n == 1){
cout << "one";
}else if ( n == 2 ){
cout << "two";
}else if ( n == 3 ){
cout << "three";
}else if ( n == 4 ){
cout << "four";
}else if ( n == 5 ){
cout << "five";
}else if ( n == 6 ){
cout << "six";
}else if ( n == 7 ){
cout << "seven";
}else if ( n == 8 ){
cout << "eight";
}else if ( n == 9 ){
cout << "nine";
}else{
cout << "Greater than 9";
}
// Write your code here
return 0;
}
4. You will be given two positive integers, and (), separated by a newline.
Sol:
int main() {
// Complete the code.
int a,b;
cin >> a >> b;
for (a;a<=b;a++){
int n =a;
if (a>9){
if (a%2==0){
cout <<"even";
}else{
cout<<"odd";
}
}else if ( n == 1){
cout << "one";
}else if ( n == 2 ){
cout << "two";
}else if ( n == 3 ){
cout << "three";
}else if ( n == 4 ){
cout << "four";
}else if ( n == 5 ){
cout << "five";
}else if ( n == 6 ){
cout << "six";
}else if ( n == 7 ){
cout << "seven";
}else if ( n == 8 ){
cout << "eight";
}else if ( n == 9 ){
cout << "nine";
}
cout<<endl;
}
return 0;
}
5. Write a function int max_of_four(int a, int b, int c, int d) which returns the maximum of the four arguments it receives.
Sol:
#include <iostream>
#include <cstdio>
using namespace std;
int max_of_four(int a,int b,int c, int d){
if(a>=b && a >= c && a >=d){
return a;
}else if( b >= c && b >=d) {
return b;
}else if( c >=d) {
return c;
}else{
return d;
}
}
int main() {
int a, b, c, d;
scanf("%d %d %d %d", &a, &b, &c, &d);
int ans = max_of_four(a, b, c, d);
printf("%d", ans);
return 0;
}
6. You will be given an array of integers and you have to print the integers in the reverse order.
Sol:
int main() {
int ar[1000],n;
cin>>n;
for(int i=0;i<n;i++){
cin >> ar[i];
}
for(int i=n-1;i>=0;i--){
cout << ar[i]<<" ";
}
return 0;
}
7. Consider an -element array, , where each index in the array contains a reference to an array of integers (where the value of varies from array to array). See the Explanation section below for a diagram. Given , you must answer queries. Each query is in the format i j, where denotes an index in array and denotes an index in the array located at . For each query, find and print the value of element in the array at location on a new line.
Sol:
int main() {
int n, c, arl, t, tt=0, x, y;
vector<int> mar;
vector<int> uvw;
scanf("%d %d",&n,&c);
uvw.push_back(0);
for (int ii=0;ii<n;ii++){
cin>>arl;
tt = tt+arl;
uvw.push_back(tt);
for (int kk=0;kk<arl;kk++){
cin>>t;
mar.push_back(t);
}
}
for (int ii=0;ii<c;ii++){
cin>>x>>y;
cout<<mar.at(uvw.at(x)+y)<<endl;
}
return 0;
}
8. Complete the update function in the editor below.
update has the following parameters:
int *a: an integer
int *b: an integer.
Sol:
void update(int *a,int *b) {
int t,x;
t=*a+*b;
x=*a-*b;
*b=x;
if (x<0)
{
*b=-x;}
*a=t;
}
9. Complete the parseInts function in the editor below.
parseInts has the following parameters:
string str: a string of comma separated integer.
Sol:
vector<int> parseInts(string str) {
// Complete this function
vector<int> vi;
string s="";
double t=0,x=0,i=0,l=0;
for(char m:str){
if(m==','){
t++;
}
l++;
}
while(x<=t && i<l){
if(str[i]==','){
if(s!="") {
vi.push_back(stoi(s));
}
s="";
i++;
x++;
}else{
s=s+str[i];
i++;
}
}
if(s!="") {
vi.push_back(stoi(s));
}
return vi;
}
10. You are given two strings, and , separated by a new line. Each string will consist of lower case Latin characters ('a'-'z').
Sol:
vector<int> parseInts(string str) {
// Complete this function
vector<int> vi;
string s="";
double t=0,x=0,i=0,l=0;
for(char m:str){
if(m==','){
t++;
}
l++;
}
while(x<=t && i<l){
if(str[i]==','){
if(s!="") {
vi.push_back(stoi(s));
}
s="";
i++;
x++;
}else{
s=s+str[i];
i++;
}
}
if(s!="") {
vi.push_back(stoi(s));
}
return vi;
}
11. Kristen is a contender for valedictorian of her high school. She wants to know how many students (if any) have scored higher than her in the exams given during this semester. Create a class named with the following specifications:
An instance variable named to hold a student's exam scores.
A void input() function that reads integers and saves them to .
An int calculateTotalScore() function that returns the sum of the student's scores..
Sol:
class Student{
public:
int scores[5];
void input(){
for(int t=0;t<5;t++){
cin>>scores[t];
}
}
int calculateTotalScore(){
int tt=0;
for(int t=0;t<5;t++){
// 7411771025;
tt+=scores[t];
}
return tt;
}
};
12. You are given integers.Sort the integers and print the sorted order.Store the integers in a vector.Vectors are sequence containers representing arrays that can change in size.
Sol:
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
vector<int> vx;
int n,m;
cin>>n;
for(int ii=0;ii<n;ii++){
cin>>m;
vx.push_back(m);
}
sort(vx.begin(),vx.end());
for(int ii=0;ii<n;ii++){
cout<<vx[ii]<<" ";
}
return 0;
}
13. The first line of the input contains an integer .The next line contains space separated integers(1-based index).The third line contains a single integer ,denoting the position of an element that should be removed from the vector. The fourth line contains two integers and denoting the range that should be erased from the vector inclusive of a and exclusive of b.
Sol:
int main() {
vector <int> vi;
int x, y, z, o, q;
cin>>x;
for(int ii=0;ii<x;ii++){
cin>>y;
vi.push_back(y);
}
cin>>z;
vi.erase(vi.begin()+z-1);
cin>>o>>q;
vi.erase(vi.begin()+o-1,vi.begin()+q-1);
cout<<(x-(q-o))-1<<endl;
for(int i=0;i<(x-(q-o))-1;i++){
cout<<vi[i]<<" ";
}
return 0;
}
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