如何进行时间分析
How to Time Analysis
我不得不为一个函数编写代码,该函数使用循环来计算从 1 到 n 的所有整数的总和。
我还需要做一个时间分析,把每个基本操作(比如赋值和++)算作一个操作。我需要帮助来理解如何计算每个基本操作。 int computeSume(int n) 是不是一步,也就是C1? for循环是多步?
请帮忙解释一下。谢谢。
#include <iostream>
using namespace std;
//Create a function that uses a loop to compute the sum of all integers from 1 to n.
//Create Function
int computeSum(int n)
{
//create sum variable
int sum = 0;
//create for loop
// i must be <= to n so it will count all integers and not cut the last int off.
for (int i = 0; i <= n; i++)
{
sum = sum + i;
}
//return statement
return sum;
}
//Main Method
int main()
{
//Input varibale
int n;
//Create user prompt
cout << "Enter a value: " << endl;
cin >> n;
cout << "The sum of all integer from 1 to " << n << " is " << computeSum(n) << "." << endl;
return 0;
}
看看这个和平,我把必要的信息都注释掉了,供大家参考。
#include <iostream>
using namespace std;
int computeSum(int n)
{
int sum = 0; // One unit time.
for (int i = 0; i <= n; i++) // Condition will be checked N+1 times so n+1 unit time.
{
sum = sum + i;
}
return sum; // One unit time.
// Total units of time is N+3, which is nothing but O(N).
}
//Main Method
int main()
{
int n; // Declaring a variable is one unit time.
cout << "Enter a value: " << endl; // One unit time.
cin >> n; // Depends how much time you take to enter value, But for simplicity taking as 1 unit.
cout << "The sum of all integer from 1 to " << n << " is " << computeSum(n) << "." << endl; // It could have been taken only a simple statement with one unit time.
// But computeSum(n) is a function so, we first analyse it's time.
return 0; // one unit.
}
我不得不为一个函数编写代码,该函数使用循环来计算从 1 到 n 的所有整数的总和。 我还需要做一个时间分析,把每个基本操作(比如赋值和++)算作一个操作。我需要帮助来理解如何计算每个基本操作。 int computeSume(int n) 是不是一步,也就是C1? for循环是多步? 请帮忙解释一下。谢谢。
#include <iostream>
using namespace std;
//Create a function that uses a loop to compute the sum of all integers from 1 to n.
//Create Function
int computeSum(int n)
{
//create sum variable
int sum = 0;
//create for loop
// i must be <= to n so it will count all integers and not cut the last int off.
for (int i = 0; i <= n; i++)
{
sum = sum + i;
}
//return statement
return sum;
}
//Main Method
int main()
{
//Input varibale
int n;
//Create user prompt
cout << "Enter a value: " << endl;
cin >> n;
cout << "The sum of all integer from 1 to " << n << " is " << computeSum(n) << "." << endl;
return 0;
}
看看这个和平,我把必要的信息都注释掉了,供大家参考。
#include <iostream>
using namespace std;
int computeSum(int n)
{
int sum = 0; // One unit time.
for (int i = 0; i <= n; i++) // Condition will be checked N+1 times so n+1 unit time.
{
sum = sum + i;
}
return sum; // One unit time.
// Total units of time is N+3, which is nothing but O(N).
}
//Main Method
int main()
{
int n; // Declaring a variable is one unit time.
cout << "Enter a value: " << endl; // One unit time.
cin >> n; // Depends how much time you take to enter value, But for simplicity taking as 1 unit.
cout << "The sum of all integer from 1 to " << n << " is " << computeSum(n) << "." << endl; // It could have been taken only a simple statement with one unit time.
// But computeSum(n) is a function so, we first analyse it's time.
return 0; // one unit.
}