C language programming implementation to calculate the effect after one year of making a little progress every day

Description of the question

The original base number is 1. If you study hard, your ability value will increase by 1% compared to the previous day, and if you let it go, your ability value will decrease by 1% compared to the previous day. What is the difference in effect after 1 year (365 days)?

Question analysis

The original base number is 1. If you work hard for one day, you will improve by 1%, and the effect will be 1*(1+0.01). If you work hard for two days, you will improve by 1% on the basis of the previous day, and the result will be 1*(1+0.01)*(1+0.01). The upward force every day after one year is (1+0.01) raised to the 365th power.

The opposite force one year later is (1-0.01) raised to the 365th power. Let’s calculate and see the difference in results.

Code

Method 1: Use for loop to implement

#include<stdio.h>
int main()
{
	int i;
	float up=1.0,down=1.0;
	for(i=1;i<=365;i++)
	{
		up=up*(1+0.01); 
		down=down*(1-0.01);
	}
	printf("每天进步一点点一年之后%5.2f\n",up);
	printf("每天退步一点点一年之后%5.2f\n",down);
	return 0;
 } 

operation result

 Method 2: while statement implementation

#include<stdio.h>
int main()
{
	int i=1;
	float up=1.0,down=1.0;
	while(i<=365)
	{
		up=up*(1+0.01); 
		down=down*(1-0.01);
		i++;
	}
	printf("每天进步一点点一年之后%5.2f\n",up);
	printf("每天退步一点点一年之后%5.2f\n",down);
	return 0;
 } 

Method 3: do-while statement implementation

#include<stdio.h>
int main()
{
	int i=1;
	float up=1.0,down=1.0;
	do
	{
		up=up*(1+0.01); 
		down=down*(1-0.01);
		i++;
	}while(i<=365);
	printf("每天进步一点点一年之后%5.2f\n",up);
	printf("每天退步一点点一年之后%5.2f\n",down);
	return 0;
 } 

Method 4:pow()

The pow() function is used to find the y power (power) of x. x, y and function values ​​are actually double types. Its prototype in use is: double pow(double x, double y); Use the pow function A header file must be added: #include<math.h>

#include<stdio.h>
#include<math.h>
int main()
{
	int i=1;
	float up=1.0,down=1.0;
	up=pow(1+0.01,365);
	down=pow(1-0.01,365);
	printf("每天进步一点点一年之后%5.2f\n",up);
	printf("每天退步一点点一年之后%5.2f\n",down);
	return 0;
 } 

think

How do you feel about the difference between making a little progress and slacking off a little?