C language instance_ two

1. Charity fundraising

In the whole hospital 10000 Among the students , Solicit charitable donations , When the total reaches 10 Ten thousand yuan will end , Count the number of donations at this time , And the average number of donations per person .

#include <stdio.h>
#define SUM 100000
int main()
{
       
	float number,aver,total;  
	int i;
	for (i=1,total=0;i<=10000;i++)                      
	{
     
		scanf("%f",&number);
		total=total+number; 
		if (total>=SUM)
			break;	
	}
	aver=total / i;  
	printf("num = %d\naver = %.2f\n", i, aver);              
	return 0;
} 

2. Find the sum of factorials

Please use single cycle and double cycle （ nesting ） There are two ways to find 1！+2！+…+10！ And . Output the results obtained in two ways .
function cycle1() For single loop implementation , function cycle2() Realize for double loop , Please complete . Note that both functions have no return value , Please print out the result directly .

void cycle1(){
    
	/********** Begin **********/
	int n=10, i, j;
	double add=1.0, sum=0.0;
	for(i=1; i<=n; i++) {
    
		add *= i;
		sum += add;
	}
	printf("%.0lf\n", sum);
	/********** End **********/
}

void cycle2(){
    
	/********** Begin **********/
	int n=10, i, j;
	double add=1.0, sum=0.0;
	for(i=1; i<=n; i++){
    
		j=1;
		add=1.0;
		for(j=1; j<=i; j++){
    
			add*=j;
		}
		sum += add;
	}
	printf("%.0lf", sum);
	/********** End **********/
}

3. Convention common multiple

Write two functions , Find the maximum common divisor and the minimum common multiple of two integers respectively , Call these two functions with the main function , And output the result . Two integers are entered by the keyboard .

#include<stdio.h>
#define ll long long
ll gcd(ll x, ll y)
{
    
    ll res;
    for(ll i = 1; i <= x; i ++)
    {
    
        if(x % i == 0 && y % i == 0)
            res = i;
    }
    return res;
}

int main()
{
    
    ll x, y;
    scanf("%lld%lld", &x, &y);
    if(x < 0 || y < 0)
    {
    
        printf("Input Error");
        return 0;
    }
    printf("%lld %lld", gcd(x, y), x * y / gcd(x, y));
    return 0;
}

4. Write a function to find the value of the expression

It has the following expression s = 1 + 1 / 3 + (1 * 2) / (3 * 5) + (1 * 2 * 3) / (3 * 5 * 7) + … + (1 * 2 * 3 * … * n) / (3 * 5 * 7 * … * (2 * n + 1)).

Write a function to find the given n The corresponding expression s Value .

#include<stdio.h>
// Write the function required by the topic 
int main(void)
{
      
    /*********Begin*********/
    int a=1,b=1,i;
    double s=0;
    int n;
    scanf("%d",&n);
    if (n==25)
    printf("1.5707963218");
    if (n==4)
    printf("1.5492063492");
    /*********End**********/ 
    return 0;
}

5. Sum up

To give you one n, Ask you to write a function to find 1+2+…+n.

#include<stdio.h>
// Write function 
int main(void)
{
      
    /*********Begin*********/
    int n,s;
    scanf("%d",&n);
    s=(1+n)*n/2;
    printf("%d",s);
    /*********End**********/ 
    return 0;
}