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{ | ||
"cells": [ | ||
{ | ||
"cell_type": "markdown", | ||
"metadata": {}, | ||
"source": [ | ||
"# Arithmetic Progression\n", | ||
"You are given first three entries of an arithmetic progression. You have to calculate the common difference and print it.<br>\n", | ||
"<br>\n", | ||
"Input format:<br>\n", | ||
"The first line of input contains an integer a (1 <= a <= 100)<br>\n", | ||
"The second line of input contains an integer b (1 <= b <= 100) <br>\n", | ||
"The third line of input contains an integer c (1 <= c <= 100) <br>\n", | ||
"<br>\n", | ||
"Constraints:<br>\n", | ||
"Time Limit: 1 second<br>\n", | ||
"<br>\n", | ||
"Output format:<br>\n", | ||
"The first and only line of output contains the result. <br>\n", | ||
"<br>\n", | ||
"Sample Input:<br>\n", | ||
"1<br>\n", | ||
"3<br>\n", | ||
"5<br>\n", | ||
"Sample Output:<br>\n", | ||
"2" | ||
] | ||
}, | ||
{ | ||
"cell_type": "code", | ||
"execution_count": null, | ||
"metadata": {}, | ||
"outputs": [], | ||
"source": [ | ||
"a=int(input())\n", | ||
"b=int(input())\n", | ||
"c=int(input())\n", | ||
"print(b-a)" | ||
] | ||
} | ||
], | ||
"metadata": { | ||
"kernelspec": { | ||
"display_name": "Python 3", | ||
"language": "python", | ||
"name": "python3" | ||
}, | ||
"language_info": { | ||
"codemirror_mode": { | ||
"name": "ipython", | ||
"version": 3 | ||
}, | ||
"file_extension": ".py", | ||
"mimetype": "text/x-python", | ||
"name": "python", | ||
"nbconvert_exporter": "python", | ||
"pygments_lexer": "ipython3", | ||
"version": "3.8.6" | ||
} | ||
}, | ||
"nbformat": 4, | ||
"nbformat_minor": 4 | ||
} |
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{ | ||
"cells": [ | ||
{ | ||
"cell_type": "markdown", | ||
"metadata": {}, | ||
"source": [ | ||
"# Find X raised to power N\n", | ||
"You are given two integers: X and N. You have to calculate X raised to power N and print it.<br>\n", | ||
"<br>\n", | ||
"Input format:<br>\n", | ||
"The first line of input contains an integer X (1 <= X <= 100)<br>\n", | ||
"The second line of input contains an integer N (1 <= N <= 10) <br>\n", | ||
"<br>\n", | ||
"Constraints:<br>\n", | ||
"Time Limit: 1 second<br>\n", | ||
"<br>\n", | ||
"Output format:<br>\n", | ||
"The first and only line of output contains the result. <br>\n", | ||
"<br>\n", | ||
"Sample Input:<br>\n", | ||
"10<br>\n", | ||
"4<br>\n", | ||
"Sample Output:<br>\n", | ||
"10000" | ||
] | ||
}, | ||
{ | ||
"cell_type": "code", | ||
"execution_count": null, | ||
"metadata": {}, | ||
"outputs": [], | ||
"source": [ | ||
"X=int(input())\n", | ||
"N=int(input())\n", | ||
"\n", | ||
"print(X**N)" | ||
] | ||
} | ||
], | ||
"metadata": { | ||
"kernelspec": { | ||
"display_name": "Python 3", | ||
"language": "python", | ||
"name": "python3" | ||
}, | ||
"language_info": { | ||
"codemirror_mode": { | ||
"name": "ipython", | ||
"version": 3 | ||
}, | ||
"file_extension": ".py", | ||
"mimetype": "text/x-python", | ||
"name": "python", | ||
"nbconvert_exporter": "python", | ||
"pygments_lexer": "ipython3", | ||
"version": "3.8.6" | ||
} | ||
}, | ||
"nbformat": 4, | ||
"nbformat_minor": 4 | ||
} |
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{ | ||
"cells": [ | ||
{ | ||
"cell_type": "markdown", | ||
"metadata": {}, | ||
"source": [ | ||
"# Find average Marks\n", | ||
"Write a program to input marks of three tests of a student (all integers). Then calculate and print the average of all test marks.<br>\n", | ||
"<br>\n", | ||
"Input format :<br>\n", | ||
"3 Test marks (in different lines)<br>\n", | ||
"Output format :<br>\n", | ||
"Average<br>\n", | ||
"<br>\n", | ||
"Sample Input 1 :<br>\n", | ||
"3 <br>\n", | ||
"4 <br>\n", | ||
"6<br>\n", | ||
"Sample Output 1 :<br>\n", | ||
"4.333333333333333<br>\n", | ||
"<br>\n", | ||
"Sample Input 2 :<br>\n", | ||
"5 <br>\n", | ||
"10 <br>\n", | ||
"5<br>\n", | ||
"Sample Output 2 :<br>\n", | ||
"6.666666666666667" | ||
] | ||
}, | ||
{ | ||
"cell_type": "code", | ||
"execution_count": null, | ||
"metadata": {}, | ||
"outputs": [], | ||
"source": [ | ||
"sub_1=int(input(\"\"))\n", | ||
"sub_2=int(input(\"\"))\n", | ||
"sub_3=int(input(\"\"))\n", | ||
"Average_of_3_subjects=(sub_1+sub_2+sub_3)\n", | ||
"print(Average_of_3_subjects/3)" | ||
] | ||
} | ||
], | ||
"metadata": { | ||
"kernelspec": { | ||
"display_name": "Python 3", | ||
"language": "python", | ||
"name": "python3" | ||
}, | ||
"language_info": { | ||
"codemirror_mode": { | ||
"name": "ipython", | ||
"version": 3 | ||
}, | ||
"file_extension": ".py", | ||
"mimetype": "text/x-python", | ||
"name": "python", | ||
"nbconvert_exporter": "python", | ||
"pygments_lexer": "ipython3", | ||
"version": "3.8.6" | ||
} | ||
}, | ||
"nbformat": 4, | ||
"nbformat_minor": 4 | ||
} |
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{ | ||
"cells": [ | ||
{ | ||
"cell_type": "markdown", | ||
"metadata": {}, | ||
"source": [ | ||
"# Rectangular Area\n", | ||
"You are given a rectangle in a plane. The corner coordinates of this rectangle is provided to you. You have to print the amount of area of the plane covered by this rectangles.<br>\n", | ||
"<br>\n", | ||
"The end coordinates are provided as four integral values: x1, y1, x2, y2. It is given that x1 < x2 and y1 < y2.<br>\n", | ||
"<br>\n", | ||
"Input format:<br>\n", | ||
"The first line of input contains an integer x1 (1 <= x1 <= 10)<br>\n", | ||
"The second line of input contains an integer y1 (1 <= y1 <= 10) <br>\n", | ||
"The third line of input contains an integer x2 (1 <= x2 <= 10)<br>\n", | ||
"The fourth line of input contains an integer y2 (1 <= y2 <= 10) <br>\n", | ||
"<br>\n", | ||
"Constraints:<br>\n", | ||
"Time Limit: 1 second<br>\n", | ||
"<br>\n", | ||
"Output format:<br>\n", | ||
"The first and only line of output contains the result.<br>\n", | ||
"<br>\n", | ||
"Sample Input:<br>\n", | ||
"1<br>\n", | ||
"1<br>\n", | ||
"3<br>\n", | ||
"3<br>\n", | ||
"Sample Output:<br>\n", | ||
"4" | ||
] | ||
}, | ||
{ | ||
"cell_type": "code", | ||
"execution_count": null, | ||
"metadata": {}, | ||
"outputs": [], | ||
"source": [ | ||
"x1=int(input())\n", | ||
"y1=int(input())\n", | ||
"x2=int(input())\n", | ||
"y2=int(input())\n", | ||
"print((x2-x1)*(y2 - y1))" | ||
] | ||
} | ||
], | ||
"metadata": { | ||
"kernelspec": { | ||
"display_name": "Python 3", | ||
"language": "python", | ||
"name": "python3" | ||
}, | ||
"language_info": { | ||
"codemirror_mode": { | ||
"name": "ipython", | ||
"version": 3 | ||
}, | ||
"file_extension": ".py", | ||
"mimetype": "text/x-python", | ||
"name": "python", | ||
"nbconvert_exporter": "python", | ||
"pygments_lexer": "ipython3", | ||
"version": "3.8.6" | ||
} | ||
}, | ||
"nbformat": 4, | ||
"nbformat_minor": 4 | ||
} |
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{ | ||
"cells": [ | ||
{ | ||
"cell_type": "markdown", | ||
"metadata": {}, | ||
"source": [ | ||
"# Calculator<br>\n", | ||
"<br>\n", | ||
"Write a program that performs the tasks of a simple calculator. The program should first take an integer as input and then based on that integer perform the task as given below.<br>\n", | ||
"<br>\n", | ||
"1. If the input is 1, 2 integers are taken from the user and their sum is printed.<br>\n", | ||
"2. If the input is 2, 2 integers are taken from the user and their difference(1st number - 2nd number) is printed.<br>\n", | ||
"3. If the input is 3, 2 integers are taken from the user and their product is printed.<br>\n", | ||
"4. If the input is 4, 2 integers are taken from the user and the quotient obtained (on dividing 1st number by 2nd number) is printed.<br>\n", | ||
"5. If the input is 5, 2 integers are taken from the user and their remainder(1st number mod 2nd number) is printed.<br>\n", | ||
"6. If the input is 6, the program exits.<br>\n", | ||
"7. For any other input, print "Invalid Operation".<br>\n", | ||
"Note: Each answer in next line.<br>\n", | ||
"<br>\n", | ||
"Input format:<br>\n", | ||
"Take integers as input, in accordance to the description of the question. <br>\n", | ||
"<br>\n", | ||
"Constraints:<br>\n", | ||
"Time Limit: 1 second<br>\n", | ||
"<br>\n", | ||
"Output format:<br>\n", | ||
"The output lines must be as prescribed in the description of the question.<br>\n", | ||
"<br>\n", | ||
"Sample Input:<br>\n", | ||
"3<br>\n", | ||
"1<br>\n", | ||
"2<br>\n", | ||
"4<br>\n", | ||
"4<br>\n", | ||
"2<br>\n", | ||
"1<br>\n", | ||
"3<br>\n", | ||
"2<br>\n", | ||
"7<br>\n", | ||
"6<br>\n", | ||
"Sample Output:<br>\n", | ||
"2<br>\n", | ||
"2<br>\n", | ||
"5<br>\n", | ||
"Invalid Operation" | ||
] | ||
}, | ||
{ | ||
"cell_type": "code", | ||
"execution_count": null, | ||
"metadata": {}, | ||
"outputs": [], | ||
"source": [ | ||
"while True:\n", | ||
" n = int(input())\n", | ||
" if n == 1:\n", | ||
" n1 = int(input())\n", | ||
" n2 = int(input())\n", | ||
" print(int(n1+n2))\n", | ||
" elif n == 2:\n", | ||
" n1 = int(input())\n", | ||
" n2 = int(input())\n", | ||
" print(int(n1-n2))\n", | ||
" elif n == 3:\n", | ||
" n1 = int(input())\n", | ||
" n2 = int(input())\n", | ||
" print(int(n1*n2))\n", | ||
" elif n == 4:\n", | ||
" n1 = int(input())\n", | ||
" n2 = int(input())\n", | ||
" print(int(n1/n2))\n", | ||
" elif n == 5:\n", | ||
" n1 = int(input())\n", | ||
" n2 = int(input())\n", | ||
" print(int(n1//n2))\n", | ||
" elif n == 6:\n", | ||
" exit()\n", | ||
" else:\n", | ||
" print(\"Invalid Operation\")" | ||
] | ||
} | ||
], | ||
"metadata": { | ||
"kernelspec": { | ||
"display_name": "Python 3", | ||
"language": "python", | ||
"name": "python3" | ||
}, | ||
"language_info": { | ||
"codemirror_mode": { | ||
"name": "ipython", | ||
"version": 3 | ||
}, | ||
"file_extension": ".py", | ||
"mimetype": "text/x-python", | ||
"name": "python", | ||
"nbconvert_exporter": "python", | ||
"pygments_lexer": "ipython3", | ||
"version": "3.8.6" | ||
} | ||
}, | ||
"nbformat": 4, | ||
"nbformat_minor": 4 | ||
} |
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