实验4

任务1：一元二次方程的根不能设置成以函数返回值的方式返回给主调函数，因为函数返回值具有唯一性，而一元二次方程的根有两个。

#include <stdio.h>
long long fac(int n);
int main() 
{
    int i,n;
    printf("Enter n: ");
    scanf("%d", &n);
    for(i=1; i<=n; ++i) 
        printf("%d! = %lld\n", i, fac(i)); 
    return 0;
}
long long fac(int n) 
{
    static long long p = 1;
    p = p*n;
    
    return p;
}

#include <stdio.h>
long long fac(int n);
int main() 
{
    int i,n;
    printf("Enter n: ");
    scanf("%d", &n);
    for(i=1; i<=n; ++i) 
        printf("%d! = %lld\n", i, fac(i)); 
    return 0;
}
long long fac(int n) 
{
    static long long p = 1;
    printf("p=%11d\n",p);
    p = p*n;
    
    return p;
}

#include<stdio.h>
int func(int, int);
int main() 
{
    int k=4,m=1,p1,p2;
    p1 = func(k,m) ; 
    p2 = func(k,m) ;
    printf("%d,%d\n",p1,p2) ;
    return 0;    
}
int func(int a,int b) 
{   
    static int m=0,i=2;
    i += m+1;
        m = i+a+b;  
        return (m);
}

#include <stdio.h>
#define N 1000
int fun(int n,int m,int bb[N]) 
{
    int i,j,k=0,flag;
    for(j=n;j<=m;j++) 
    {
        flag=1;   
        for(i=2;i<j;i++)
            if(j%i==0) 
            {  
               flag=0;
               break;
            }
        if(j==i) 
           bb[k++]=j;
    }
    return k;
}
int main()
{
    int n=0,m=0,i,k,bb[N];
    scanf("%d",&n);
    scanf("%d",&m);
    for(i=0;i<m-n;i++)
        bb[i]=0;
    k=fun(n,m,bb); 
    for(i=0;i<k;i++)
        printf("%4d",bb[i]);
    return 0;
}

#include <stdio.h>
long long fun(int n);
int main() 
{
    int n;
    long long f;
    while(scanf("%d", &n) != EOF) 
    {
        f = fun(n);
        printf("n = %d, f = %lld\n", n, f);
    }
    return 0;
}
long long fun(int n)
{
    long long f;
    if(n==0)
        return 0;
    else
        f=2*fun(n-1)+1;
    return f;
}

#include <stdio.h>
void draw(int n, char symbol);
#include <stdio.h> 
int main() 
{
    int n, symbol;
    while(scanf("%d %c", &n, &symbol) != EOF) 
    {
        draw(n, symbol);
        printf("\n");
    }
    return 0;
} 
void draw(int n,char symbol)
{
    int line,j;
    for(line=0;line<n;line++)
    {
        for(j=0;j<n-line-1;j++)
        {
            printf(" ");
        }
        for(j=0;j<=2*line;j++)
        {
            printf("%c",symbol);
        }
        printf("\n");
    }
}