Problem 1. Kitchen Timings
The working hours of Chef’s kitchen are from X pm to Y pm (1≤X<Y≤12).
Find the number of hours Chef works.
Input Format
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of two space-separated integers X and Y — the starting and ending time of working hours respectively.
Output Format
For each test case, output on a new line, the number of hours Chef works.
Constraints
- 1 ≤ T ≤ 100
- 1 ≤ X < Y ≤ 12
In C, I originally used this like nested while-loop but it would just print out a bunch of one’s. Changed the second while-loop to an if-statement and now I get the correct answer. Also, most of these problems I don’t incorporate the “constraints” that are list on the original problems and just write the code. Except this one I incorporated the constraints into the code.
#include <stdio.h>
int main(void)
{
int t;
scanf("%d", &t);
while(t--)
{
int x, y;
scanf("%d %d", &x, &y);
if (1 <= x && x < y && y <= 12)
{
printf("%d\n", y - x);
}
}
return 0;
}
for t in range(int(input())):
x, y = map(int, input().split())
if (1 <= x and x < y and y <= 12)
print(y - x)
Problem
In Chefland, precipitation is measured using a rain gauge in millimeter per hour.
Chef categorizes rainfall as:
LIGHT
, if rainfall is less than 3 millimetre per hour.MODERATE
, if rainfall is greater than equal to 3 millimeter per hour and less than 7 millimeter per hour.HEAVY
if rainfall is greater than equal to 7 millimeter per hour.
Given that it rains at X millimeter per hour on a day, find whether the rain is LIGHT
, MODERATE
, or HEAVY
.
Input Format
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of a single integer X — the rate of rainfall in millimeter per hour.
Output Format
For each test case, output on a new line, whether the rain is LIGHT
, MODERATE
, or HEAVY
.
You may print each character in lowercase or uppercase. For example, LIGHT
, light
, Light
, and liGHT
, are all identical.
Constraints
- 1 ≤ T ≤ 20
- 1 ≤ X ≤ 20
#include <stdio.h>
int main(void)
{
int t;
scanf("%d", &t);
while(t--)
{
int x;
scanf("%d", &x);
if(x < 3)
printf("Light\n");
else if(3 <= x && x < 7)
printf("Moderate\n");
else
printf("Heavy\n");
}
return 0;
}
for t in range(int(input())):
x = int(input())
if(x < 3):
print("Light")
elif(3 <= x and x < 7):
print("Moderate")
else:
print("Heavy")
Problem 3. Spice Level
Each item in Chef’s menu is assigned a spice level from 11 to 1010. Based on the spice level, the item is categorized as:
MILD
: If the spice level is less than 4.MEDIUM
: If the spice level is greater than equal to 4 but less than 7.HOT
: If the spice level is greater than equal to 7.
Given that the spice level of an item is X, find the category it lies in.
Input Format
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of an integer X — the spice level of the item.
Output Format
For each test case, output on a new line, the category that the item lies in.
You may print each character in uppercase or lowercase. For example, HOT
, hot
, Hot
, and hOT
are all considered the same.
Constraints
- 1 ≤ T ≤ 1000
- 1 ≤ X ≤ 10
#include <stdio.h>
int main(void)
{
int t;
scanf("%d", &t);
while(t--)
{
int x;
scanf("%d", &x);
if(x < 4)
printf("Mild\n");
else if(4 <= x && x < 7)
printf("Medium\n");
else
printf("Hot\n");
}
return 0;
}
for t in range(int(input())):
x = int(input())
if(x < 4):
print("Mild")
elif(4 <= x and x < 7):
print("Medium")
else:
print("Hot")
Problem 4. Scalene Triangle
Given A,B, and C as the sides of a triangle, find whether the triangle is scalene.
Note:
- A triangle is said to be scalene if all three sides of the triangle are distinct.
- It is guaranteed that the sides represent a valid triangle.
Input Format
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of three space-separated integers A, B, and C — the length of the three sides of the triangle.1
Output Format
For each test case, output on a new line, YES
, if the triangle is scalene, and NO
otherwise.
You may print each character of the string in uppercase or lowercase. For example, YES
, yes
, Yes
, and yEs
are all considered identical.
Constraints
- 1 ≤ T ≤ 100
- 1 ≤ A ≤B ≤ C ≤ 10
- C < (A + B)
#include <stdio.h>
int main(void)
{
int t;
scanf("%d", &t);
while(t--)
{
int a, b, c;
scanf("%d %d %d", &a, &b, &c);
if(a != b && b != c && c < a + b)
printf("Yes\n");
else
printf("No\n");
}
return 0;
}
for t in range(int(input())):
a, b, c = map(int, input().split())
if(a != b and b != c and c < a + b):
print("Yes")
else:
print("No")
# another way to write this
for t in range(int(input())):
a, b, c = map(int,input().split())
if a == b or b == c or a == c: print("NO")
else: print("YES")