// Name: Anita Blake Partner: Ted Forrester
// ID: 014916 ID: 642708
// Sec: Yan T 2:30 Sec: Artemov R 1:25
// Date: Feb 10, 2000
// Project 2: Loops Ahoy!
import java.text.DecimalFormat;
// 1. Compute miles per gallon of a trip. Assume no input errors.
// Input: gas used, start odometer reading, end odometer reading
public class Project2_1 {
public static void main(String[] args) {
double gas; // gallons of gas used
int start_mile; // beginning odometer reading in mile
int end_mile; // end of trip odometer reading in mile
// Initialize Text object in to read from standard input.
TokenReader in = new TokenReader(System.in);
System.out.println("1. Miles per gallon");
System.out.println("Enter gallons of gas used and odometer readings at");
System.out.println("start and end of trip rounded to the nearest mile");
gas = in.readDouble();
start_mile = in.readInt();
end_mile = in.readInt();
System.out.println("Gas economy for this trip is " +
(end_mile-start_mile)/gas + " miles per gallon.");
}
}
// 2. Count multiples of 5 in an input sequence.
public class Project2_2 {
public static void main(String[] args) {
// Initialize Text object in to read from standard input.
TokenReader in = new TokenReader(System.in);
System.out.println("2. Count multiples of 5 in an input sequence");
System.out.println("Enter a sequence of positive integers " +
"terminated by a non-positive integer.");
int count=0; // no. of multiples of 5 found so far
int input = in.readInt(); // item in input sequence to be processed
while (input > 0) {
if (input%5 == 0) count=count+1;
input = in.readInt();
}
System.out.println(count +" multiple(s) of 5 found " +
"before stopping signal");
}
}
// 3. Count number of sign changes in an input sequence.
public class Project2_3 {
public static void main(String[] args) {
// Initialize Text object in to read from standard input.
TokenReader in = new TokenReader(System.in);
int LIMIT = 100; // input range is -LIMIT to +LIMIT
System.out.println("3. Count number of sign changes in input sequence");
System.out.println("Enter integers between -" + LIMIT + " and +" +
LIMIT + ". End sequence with out-of-range number.");
int change = 0; // number of sign changes found so far
int current = in.readInt(); // current input value
int previous = current; // previous input value, initialize to be
// same sign as 1st input value
while ( Math.abs(current) <= LIMIT ) {
// Compare previous and current. There is a sign change if product
// is negative, or if product is zero AND not both are zeros.
if ( previous*current<=0 && !(previous==0 && current==0) )
change = change+1;
/* Another solution: the test could be that both numbers
are negative, both 0, or both positive:
if ( previous < 0 && current < 0
|| previous == 0 && current == 0
|| previous > 0 && current > 0
)
change = change+1;
*/
// Update previous and read next input
previous = current;
current = in.readInt();
}
System.out.println(change + " sign change(s) found before " +
"stopping signal");
}
}
// 4. Find two neighboring elevations with the steepest gradient.
public class Project2_4 {
public static void main(String[] args) {
int current; // current contour value
int previous; // previous contour value
int gradient; // gradient between current, previous contours
int maxGradient; // steepest gradient found so far
int elevationL; // left neighbor of steepest gradient so far
int elevationR; // right neighbor of steepest gradient so far
int stopValue=-1; // stopping value to indicate end of input
// Initialize Text object in to read from standard input.
TokenReader in = new TokenReader(System.in);
System.out.println("4. Neighboring elevations with steepest gradient");
System.out.println("Enter contour readings terminated by -1");
maxGradient = 0; // initialize to smallest possible value
elevationL = stopValue; // initialize to stopValue
elevationR = stopValue; // initialize to stopValue
current = in.readInt(); // read first input value
previous = current; // initialize to get bogus gradient of 0
while ( current != stopValue ) {
// If computed gradient greater than maxGradient,
// update maxGradient, elevationL, elevationR
gradient = Math.abs(previous-current);
if (gradient>maxGradient) {
maxGradient = gradient;
elevationL = previous;
elevationR = current;
}
// Update previous and read next input
previous = current;
current = in.readInt();
}
if (elevationL == stopValue)
System.out.println("Insufficient number of samples to have " +
"a maximum gradient");
else
System.out.println("Steepest gradient occurs between elevations "
+ elevationL + " and " + elevationR);
}
}
===============================================================================
1. Miles per gallon
Enter gallons of gas used and odometer readings at
start and end of trip rounded to the nearest mile
5 13 140
Gas economy for this trip is 25.4 miles per gallon.
2. Count multiples of 5 in an input sequence
Enter a sequence of positive integers terminated by a non-positive integer.
5 21 1 21 25 32 -1
2 multiple(s) of 5 found before stopping signal
2. Count multiples of 5 in an input sequence
Enter a sequence of positive integers terminated by a non-positive integer.
0 5 -1
0 multiple(s) of 5 found before stopping signal
2. Count multiples of 5 in an input sequence
Enter a sequence of positive integers terminated by a non-positive integer.
150 -1
1 multiple(s) of 5 found before stopping signal
3. Count number of sign changes in input sequence
Enter integers between -100 and 100. End sequence with out-of-range number.
-99 -97 12 -10 1 -2 101
4 sign change(s) found before stopping signal
3. Count number of sign changes in input sequence
Enter integers between -100 and 100. End sequence with out-of-range number.
1 -1 0 1 -1 101
4 sign change(s) found before stopping signal
3. Count number of sign changes in input sequence
Enter integers between -100 and 100. End sequence with out-of-range number.
1 -101
0 sign change(s) found before stopping signal
3. Count number of sign changes in input sequence
Enter integers between -100 and 100. End sequence with out-of-range number.
2000
0 sign change(s) found before stopping signal
4. Neighboring elevations with steepest gradient
Enter contour readings terminated by -1
52 63 63 95 100 -1
Steepest gradient 32 is between elevations 63 and 95
4. Neighboring elevations with steepest gradient
Enter contour readings terminated by -1
0 -1
Insufficient number of samples to have a maximum gradient