Type the program's output stop = int(input ( ) ) result = 0 for a in range (2): print (a, end=': ') for b in range (4): result += a + b if result > stop: print () print('-', end=' ') continue print (result, end=' ') Input 6 Output 1
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- 1. Rewrite the program segment that follows, using a for loop and do-while loop: = 1; product i = 0; %3D while (i < n) { scanf(“%d", &a); if (a != i) %3D product *= a; ++i; }module test (clk, a, b, f, g, h); input clk, a, b; output f, g, h; reg f, g, h; always@ (negedge glk) f <= #5 a| b; alwayse (clk) g = #10 f ^ b; alwayse (f or g) h = #2 f & g; endmodule Figure 1JAVA LANGUAGE PLS THANK YOU 1. Squares by CodeChum Admin Looping random numbers and manipulating those values are fun, too! Why don't we try one that returns the square of an inputted number repeatedly until the user inputs 0? Let's do this! Instructions: Using a do...while() loop, continuously scan for random integers that will be inputted by the user and print out its square, separated in each line. Once the inputted value is 0, it will still print out its square value but should then terminate the loop afterwards. Use this concept in making your loop condition. Input Multiple lines containing an integer. 2 6 0 Output Multiple lines containing an integer. 4 36 0
- module test (clk, a, b, f, g, h); input clk, a, b; output f, g, h; reg f, g, h; always@ (negedge clk) f <= #5 a | b; always@ (clk) g = #10 £ ^ b; always@ (f or g) #2 h = f & g; endmodule Figure 1LAB: Loops: Output range with increment of 10 Write a program that reads two integers as input, and outputs the first integer and subsequent increments of 10 as long as the value is less than or equal to the second integer. Ex: If the input is: -15 30 the output is: -15 -5 5 15 25 Ex: If the second integer is less than the first as in: 20 5 the output is: Second integer can't be less than the first. For coding simplicity, output a space after every integer, including the last.All programming languages have a loop capability. Looping is the program logic’s ability to repeat one or more lines of code either: A set number of times, a “for loop” (i.e. 10, 100, 500, 763 or more times) An “unknown” number of times (i.e. loop terminates if a tested condition becomes “true” or as long as the tested condition remains “false”), a “while” loop, a “do until” loop, a “do while” loop. This Lab exercise demonstrates the use of the “For Loop”, a standard loop in all programming languages. C++ standard For Loop format looks like: For (n = 1; n <= 100; n++) { Loop line 1 code; Loop line 2 code; Loop line 3 code; Loop line N code; } The loop code needs: A “loop counter variable” (n in the above example) to keep track of how many iterations this loop has run A starting value (i.e. 1 in the above example) A stop test (i.e. run as…
- Debug Program4: main() { int m,n; for(i 0; i << 5; i++) { printf("\t\t\t\t"); for(j 0; j << 5; j+++) printf("*"); printf("/n"); } Program 5: int func(int) main() { int num = 7, c; int f= func(num); printf("\n\n%d of %d\n\n", num, f); } int func(aj); { if(aj=1 aj=0) return 1; else return (aj*fun(aj-1); } 49 5040 7117 none of the aboveelearning.ibrict.edu.om EE EECP1290: Computer Programming for Engineering-CLA#1 Write a program to calculate the K value using if else condition If P value is more than 9 then K = M/2 VP+C If P value is less than or equal to 9 then K = M2 + P+C Where C is a constant with value 1.7 Sample output: Enter p value:16 Enter M value:4 p, M and K values are:16.000000 4.000000 -0.300000 Sample output: Enter p value:9 Enter M value:4 P, M and K values are:9.000000 4.000000 6.700000 >Q1-Given below is an example of an unstructured program containing an infinite loop and statements like break/continue. void main() (int a, b, I; scanf("%d %d", &a, &b) 1 = -1; while (1) } if (b 1). (b) Convert the above program into an equivalent structured program (i.e. without using break/continue and infinite loop).
- Examine the following for loops and determine the value of ires at the end of each of the loops, and also the number of times each loop executes. (a) ires = 0; for index = -11:11 ires = ires + 1; end ires = number of times (b) ires = 0; for index = 10:-2:1 if index == 6 continue; end ires = ires + index; end ires = number of times ires = 0; for index = 10:-2:1 if index == 6 (c) break; end ires = ires + index: end ires = number of times (d) ires = 0; for indexl = 10:-2:1 for index2 = 2:2:index1 if index2 == 6 break end ires = ires + index2; end end ires = number of times (outer loop) number of times (inner loop)NOT using C++ code format Write a pseudocode for the real percentage-to-numerical-grade-equivalent displaying. Only one percentage score can be entered at a time. The grading scale is as follows: example link: (pls follow this one) https://www.code4example.com/pseudocode/pseudocode-examples/main() { int a = 1, b = 2, c = 3: printf("%d", a + = (a + = 3, 5, a)) } Output of the given program will be ?