实验4

实验任务1

#include<stdio.h>
#include<math.h>
void solve(double a,double b,double c);
int main(){
    double a,b,c;
    printf("Enter a,b,c:");
    while(scanf("%lf%lf%lf",&a,&b,&c)!=EOF){
        solve(a,b,c);
        printf("Enter a,b,c:");
        
    }
    return 0;
}
void solve(double a,double b,double c){
    double x1,x2;
    double delta,real,imag;
    if(a==0)
       printf("not quadratic equation.\n");
       else{
           delta=b*b-4*a*c;
           
           if(delta>=0)
           x1=(-b+sqrt(delta)/(2*a));
           x2=(-b-sqrt(delta)/(2*a));
           printf("x1=%f+%fi,x2=%f-%fi\n",real,imag,real,imag);
       }
}

一元二次方程的根不可以设计成通过函数返回值的方式传递回主函数，函数的返回值只有一个，一元二次方程可能有两个根。

 

实验任务2

#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!=%11d\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!=%11d\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); 
}

第一次，m=8，返回m=8，p1=8;第二次，m=17,返回m=17,p2=17

实验任务3

#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<=m;i++)
        if(j%i==0){
            flag=0;
            break;
        } 
        if(i>=j)
        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;
    }

实验任务4

#include<stdio.h>
long long fun(int n);

int main(){
    int n;
    long long f;
    
    while(scanf("%d",&n)!=EOF){
        f=fun(n)-1;
        printf("n=%d,f=%d\n",n,f);
        
    }
    return 0;
}
int long long fun(int n){

    int k;
    if(n==1)
    k=2;
    else
    k=2*fun(n-1);
    return k;
    }
    

这里有问题，换了几种方式，基本都是这个结果。为啥

 

实验任务5

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