实验4

ex1

一元二次方程的根不能设计成为函数的返回值，因为函数只能返回一个值，但是一元二次方程的根可能有两个。

ex2_1

#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;
}

增加一行代码

#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=%lld\n",p);
    p = p*n;
    
    return p;
}

ex2_2

#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);
}

ex3

#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(flag==1) 
           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;
}

ex4

#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 s;
    if(n==0)
    s=0;
    else
    s=2*fun(n-1)+1;
    return s;
    
}

ex5

#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 i,j,k;
    for(i=1;i<=n;i++){
        for(j=1;j<=n-i;j++)
        printf(" ");
        for(k=1;k<2*i;k++)
        {
            printf("%c",symbol);
        }
        printf("\n");
    }
    
}