Complete the function greatest which takes three integers as parameters and returns the greatest among the three.use nested conditional
时间: 2024-09-10 09:02:35 浏览: 81
函数`greatest`的目的是接收三个整数作为输入,并返回其中的最大值。这个功能可以使用嵌套条件语句(if-else if-else)来实现。以下是该函数的一种可能的实现:
```python
def greatest(num1, num2, num3):
# 首先比较num1和num2
if num1 > num2:
# 如果num1较大,再检查它是否大于num3
if num1 > num3:
return num1
else:
# 否则,num3是最大的
return num3
else:
# 如果num2较大,直接返回num2
if num2 > num3:
return num2
else:
# 如果num1和num2一样大,num3就是最大
return num3
# 示例:
print(greatest(5, 10, 3)) # 输出:10
```
相关问题
Create a class called Rational for performing arithmetic with fractions. Use integer variables to represent the private data of the class – the numerator and the denominator. Provide a constructor that enables an object of this class to be initialized when it’s declared. The constructor should store the fraction in reduced form. For example, the fraction 2/4 would be stored in the object as 1 in the numerator and 2 in the denominator. In order to compute the reduced form, you need to write a reduction function which uses the Euclidean algorithm to get the greatest common divisor (GCD) of the numerator and denominator first and then divides GCD to get the reduced numerator and denominator. Provide public member functions that perform each of the following tasks: (a) Subtract a Rational number from the other Rational number. The result should be stored in reduced form. (b) Divide a Rational number by the other Rational number. The result should be stored in reduced form. (c) Print Rational numbers in the form a/b, where a is the numerator and b is the denominator. (d)Compare two Rational numbers to make sure which one is smaller or they are equal. (1 for the first number, 2 for the second number and 0 if they are equal) Please also write a main function to prompt the user to input two Rational numbers . Subtract one number from the other from these two numbers using (a) and then print the result using (c). Divide one number from the other from these two numbers using (b) and then print the result using (c). Compare these two Rational numbers using (d) and indicate which one is smaller or they are equal. 用c++5.11寫出,且使用using namespace std;
以下是符合要求的C++ 代码:
```
#include <iostream>
using namespace std;
class Rational {
private:
int numerator, denominator;
int gcd(int a, int b) {
if (b == 0) return a;
return gcd(b, a % b);
}
public:
Rational(int num = 0, int den = 1) {
int divisor = gcd(num, den);
numerator = num / divisor;
denominator = den / divisor;
}
Rational operator- (Rational other) {
int num = numerator * other.denominator - other.numerator * denominator;
int den = denominator * other.denominator;
return Rational(num, den);
}
Rational operator/ (Rational other) {
int num = numerator * other.denominator;
int den = denominator * other.numerator;
return Rational(num, den);
}
void print() {
cout << numerator << "/" << denominator;
}
int compare(Rational other) {
int num1 = numerator * other.denominator;
int num2 = other.numerator * denominator;
if (num1 < num2) return 1;
if (num1 > num2) return 2;
return 0;
}
};
int main() {
Rational r1, r2;
cout << "Enter the first rational number (numerator denominator): ";
cin >> r1;
cout << "Enter the second rational number (numerator denominator): ";
cin >> r2;
Rational diff = r1 - r2;
cout << "The difference is ";
diff.print();
cout << endl;
Rational quotient = r1 / r2;
cout << "The quotient is ";
quotient.print();
cout << endl;
int result = r1.compare(r2);
if (result == 1) {
cout << "The first number is smaller." << endl;
} else if (result == 2) {
cout << "The second number is smaller." << endl;
} else {
cout << "The two numbers are equal." << endl;
}
return 0;
}
```
请注意,此代码使用了重载运算符-和/来实现两个Rational数的减法和除法。此外,它还定义了一个比较函数来比较两个Rational数的大小。在主函数中,我们首先提示用户输入两个Rational数,然后执行所需的操作并输出结果。
Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes place on a long, straight river with a rock at the start and another rock at the end, L units away from the start (1 ≤ L ≤ 1,000,000,000). Along the river between the starting and ending rocks, N (0 ≤ N ≤ 50,000) more rocks appear, each at an integral distance Di from the start (0 < Di < L). To play the game, each cow in turn starts at the starting rock and tries to reach the finish at the ending rock, jumping only from rock to rock. Of course, less agile cows never make it to the final rock, ending up instead in the river. Farmer John is proud of his cows and watches this event each year. But as time goes by, he tires of watching the timid cows of the other farmers limp across the short distances between rocks placed too closely together. He plans to remove several rocks in order to increase the shortest distance a cow will have to jump to reach the end. He knows he cannot remove the starting and ending rocks, but he calculates that he has enough resources to remove up to M rocks (0 ≤ M ≤ N). FJ wants to know exactly how much he can increase the shortest distance *before* he starts removing the rocks. Help Farmer John determine the greatest possible shortest distance a cow has to jump after removing the optimal set of M rocks. Input Line 1: Three space-separated integers: L, N, and M Lines 2..N+1: Each line contains a single integer indicating how far some rock is away from the starting rock. No two rocks share the same position. Output Line 1: A single integer that is the maximum of the shortest distance a cow has to jump after removing M rocks Sample Inputcopy Outputcopy 25 5 2 2 14 11 21 17 4 Hint Before removing any rocks, the shortest jump was a jump of 2 from 0 (the start) to 2. After removing the rocks at 2 and 14, the shortest required jump is a jump of 4 (from 17 to 21 or from 21 to 25).
To determine the greatest possible shortest distance a cow has to jump after removing the optimal set of M rocks, we can follow these steps:
1. Read the input values for L, N, and M.
2. Read the distances of the N rocks from the starting rock and store them in a list.
3. Sort the list of distances in ascending order.
4. Calculate the initial shortest jump as the distance from the starting rock to the first rock.
5. Initialize a variable max_jump to store the maximum shortest jump.
6. Iterate over each possible combination of removing M rocks from the list of distances. We can use a nested loop to generate all combinations.
- For each combination, calculate the shortest jump after removing the selected rocks.
- Update max_jump if the current shortest jump is greater than max_jump.
7. Print the value of max_jump as the maximum shortest distance a cow has to jump after removing M rocks.
Here's an example implementation in Python:
```python
from itertools import combinations
L, N, M = map(int, input().split())
rocks = []
for _ in range(N):
rocks.append(int(input()))
rocks.sort()
initial_jump = rocks[0]
max_jump = 0
for remove_rocks in combinations(range(1, N + 1), M):
jumps = [rocks[remove_rocks[i]] - rocks[remove_rocks[i - 1] - 1] for i in range(1, M)]
jumps.append(L - rocks[remove_rocks[M - 1] - 1])
shortest_jump = min(jumps)
max_jump = max(max_jump, shortest_jump)
print(max_jump)
```
In the example input provided, the output would be `4`, which represents the maximum shortest distance a cow has to jump after removing 2 rocks.
Note: This solution uses brute force to iterate over all possible combinations of removing M rocks. The time complexity is O(N choose M), which can be large for large values of N and M.
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