My solutions to the Very Easy questions

Easy Questions

1. int suit, number;

for (int i = 0; i < 2; i = i + 1)
suit = rand()%4+1
number = rand()%13+1

if (number == 1)
cout << "You drew the Ace of";
if (number == 11)
cout << "You drew the Jack of";
if (number == 12
cout << "You drew the Queen of";
if (number == 13)
cout << "You drew the King of";
if (number > 1 && number < 11)
cout << "You drew the" << number << "of";

if (suit == 1)
cout << " Spades";
##etc for other suits

2. int time = 9999

int SS = time %60
int MM = time /60
time = time - SS
int MM = time %60
int HH = time/60

cout << HH << ":" << MM << ":" << SS

3. int resultA, resultB, resultC;
int coinA, coinB;
for (int i = 0; i < 1001; i = i + 1)
{coinA = rand()%2
coinB = rand()%2
if (coinA+coinB == 0)
resultA = resultA + 1;
if (coinA+coinB == 1)
resultB = resultB + 1;
if (coinA+coinB == 2)
resultC = resultC + 1;}

float percentageA, percentageB, percentageC;
percentageA = (resultA/1000)*100
percentageB = (resultB/1000)*100
percentageC = (resultC/1000)*100

cout << "The percentage of two heads was" << percentageA
cout << "The percentage of two tails was" << percentageB
cout << "The percentage of one head and one tail was" << percentageC

4. int C[14]

for (int i = 0; i < 5; i = i +1)
C[i] = A[i]

for (int i = 0; i < 10; i = i+1)
C[i+5] = B[i]

5. int money;
float interest;
int x; (years)

for (int i = 0; i < x; i = i + 1)
money = money * (interest+1)

6. int X[199];
int tempstorage;

for (int i = 0; i < 200; i = i + 1)
X[i] = rand()%11

for (int i = 0; i < 200; i = i + 1)
for (int j = 0; j < 200; j = j + 1)
if X[j] < X[i]
{tempstorage = X[i];
X[i] = X[j];
X[j] = tempstorage;}

(same for the other list, just swap the <> direction)

I'm too lazy to do the mode one.

7. int A, B, C;
float C
cin << A << B

while (A > 0)
{if (A - B < 0)
{D = A;
break;}
A = A - B
C = C + 1
}

cout << "The result of the division is" << C << "with a remainder of" << D

8. int roman;
cin << roman;

if roman == 10
cout << "X"

if roman > 5
{cout << "V";
roman = roman - 5;
if (roman == 4)
cout << "IV"
else
while (roman > 0)
{cout << "I";
roman = roman - 1;}}

if roman == 5
cout << "V"

if roman < 5
{if roman == 4
cout << "IV"
else
while (roman > 0)
{cout << "I";
roman = roman - 1;}}

9. ##num1 and num2 are the two integers to be added, num5 is the final number, the rest are for code usage

int num1, num2, num3, num5;
float num4;
num4 = num1;

##counting the number of digits by dividing by 10 until the number is a decimal
while (num4 > 1)
{num3 = num3 + 1;
num4 = num4 /10;}

num5 = num1 + (num2 * pow(10,num3))
cout << num5

10. Don't know how to do this.

11. I haven't the slightest clue how this maths.

12. for (int i = 0; i < 1000; i = i + 1)
for (int j = 0; j < 1000; j = j + 1)
if ( i != j && X[i] == X[j])
cout << "The repeated number was << cout << X[i];

follow me @realdonaldtrump or @mido9_2 (I have no tweets, this is just a watermark)