实验3 C语言分支语句、循环语句、函数综合应用编程1

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define N 5

int main() {
    int x, n;
    
    srand(time(0)); 
    
    for(n=1; n<=N; n++) {
        x = rand() % 32;
        printf("%3d", x);
    }

    printf("\n");
     
    return 0;
} 

 

 

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define N 1

int main() {
    int x, n,m;
    n=0;
    
    printf("猜猜2021年5月哪一天会是你的luck day\n开始喽，你有三次机会，猜吧（1~31）");
    
    srand(time(0)); 
     x = rand() % 32;

    while(n<3){
        scanf("%d",&m);
        if(m<x){
            if(n<2)
            printf("你猜的日期早了，luck day还没到呢\n再猜：");
            else
            printf("你猜的日期早了，luck day还没到呢");
        }
        
        else if(m>x){
            if(n<2)
            printf("你猜的日期晚了，luck day悄悄溜到前面啦\n再猜:");
            else
            printf("你猜的日期晚了，luck day悄悄溜到前面啦");    
        
        }
        
        else{
            printf("猜对了");
            return 0; 
        }
        
         n++;
        
    }
    printf("\n次数用完啦。偷偷告诉你，5月，你的luck day是%d号",x);
     
    return 0;
} 

 

 

 

 

#include <stdio.h>

int main(){
    
    long int m,n,p,q;
    p=1;
    q=0;
    
    printf("Enter a number:");
    scanf("%d",&m);
    
    while(m!=0){
        n=m%10;
        m=m/10;
        
        if(n%2!=0){
            q=q+n*p;
            p=p*10;
        }
    }
    printf("new number is:%d",q);
    
    return 0;
}

 

 

#include <math.h>
#include <stdio.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 = %.2f, x2 = %.2f\n", x1, x2);
       }
       else {
          real = -b/(2*a);
          imag = sqrt(-delta) / (2*a);
          printf("x1 = %.2f + %.2fi, x2 = %.2f - %.2fi\n", real, imag,real, imag);
        }
    }
}

 

否，最后得出的答案有两个

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

int main() {
    int n;
    double s;

    printf("Enter n(1~10): ");

    while(scanf("%d", &n) != EOF) {
        s = fun(n); 
        printf("n = %d, s= %f\n\n", n, s);
        printf("Enter n(1~10): ");
    }
    
    return 0;
}

double fun(int n) {
    double p=1,q=0,i;
     
    for(i=1;i<=n;i++){
        q=q+p;
        p=(-1)*p*(1/(i+1));
 
    }
    return q;
}

 

 

#include<stdio.h>
#include<math.h>
int isprime(int n);
int main()
{
    int i,x=0,y=0;
    for(i=101;i<=200;i++)
    {
        if(isprime(i))
            {x++;
             printf("%4d",i);
            if(x%5==0)
            printf("\n");
            y=y+1;
             }
    }
    printf("\n\n");
    printf("100~200之间的素数个数为：%d",y);
    printf("\n");
    return 0;
}
int isprime(int n)
{
    int k;
    for(k=2;k<=sqrt(n);k++)
        if(n%k==0)
            return 0;
        return 1;
}