第一章：喝汽水，阶梯电费计算，阶梯电费计算函数，个人所税，求解平方根不等式，简化求解平方根不等式，求解调和级数不等式，解不等式：d＜1+1/2-1/3+1/4+1/5-1/6+..士1/n

//喝汽水
int main()
{
    
	long m, t, x, y;
	printf("请输入人数:");
	scanf("%d", &m);
	x = m / 20;
	t = m - 20 * x;
	y = t / 3;
	t = m - 20 * x - 3 * y;
	printf("喝%ld瓶汽水，至少需要%.2f元", m, (13 * x + 2 * y) * 1.4 + t);
	return 0;
	
}

验证结果：

//阶梯电费计算
int main()
{
    
    double x,y;
    printf(" 请输入月用电量（度)：");
    scanf ("%lf", &x); 
    if( x <= 240) //1/分3档计 
    y = x * 0.49;
    else if (x <= 400) 
     y= 240 * 0.49 + (x - 240) *0.54;
    else 
     y = 240 * 0.49 +(400-240) *0.54 + (x - 400) *0.79;
    printf(" 应缴电费（元）：%9.2f\n",y);
    return 0;
}

结果：

//阶梯电费计算函数
 double f (double x)
 {
    
    double y; 
    if (x <= 240) 
       y= x * 0.49;
    else if (x <= 400) 
       y=f (240) + (x-240) *0.54; //调用自身函数
    else y=f (400) + (x-400) *0.79;
     return (y); //返回电费值了

 }
int main()
{
    
    double x;
    printf(" 请输入月用电量： ");
    scanf("%lf", &x);
    printf("应交电费 %9.2f\n",f(x));

    return 0;
}

结果：

//个人所得税
double s(double x)
{
    
    double y, c, d;
    d=0;
     c=3500+d; 
     if (x<=c) y=0; 
     else if (x-c <= 1500) 
	 	y= (x-c) *0.03;
      else if (x-c <= 4500) 
	  	y=s (c+1500) + (x-c-1500) *0.10; 
      else if (x-c <= 9000) 
	  	y=s(c+4500) + (x-c-4500) *0.20; 
      else if (x-c <= 35000)
	  	 y=s (c+9000) + (x-c-9000) *0.25; 
      else if (x-c <= 55000)
	  	 y=s (c+35000) + (x-c-35000) *0.30; 
      else if (x-c <= 80000) 
	  	y=s (c+55000) + (x-c-55000) *0.35; 
      else 
	  	y =s(c + 80000) + (x-c-80000) *0.45; 
      return (y) ;
}
int main()
{
    
    double x;
    printf(" 请输入月收入金额:");
    scanf("%lf", &x);
    printf(" 应交个人所税：%9.2f\n", s(x));

    return 0;
}

结果：

//求解平方根不等式
int main()
{
    
    long i, m; double n, s, s1;
    printf(" 请输入正数n (n>3)：");
    scanf ("%lf", &n) ; //输入任意正数 
    m=0;
    while (1)
    {
     
	 m++;
     s=0; 
     for (i=m; i<=2*m; i++) 
     s=s+sqrt (i) ; //对每一个川计算和 s 
     if (s >= n) break;
      else s1=s; //为以下注明提供依据
    }
    printf(" 不等式的解为 m >= %ld\n",m);
    printf("注：当m = %ld时， s = %.2f; 当 m = %ld时， s = %.2f\n", m-1, s1, m, s);
    return 0;
}

结果：

//简化求解平方根不等式 
int  main()
{
    
    long m;
     double n, s, s1;
     printf(" 请给入正数n (n > 3)：");
    scanf ("%lf", &n) ;//输入任意正数 
     m = 1;
     s = 1.0+ sqrt (2);
     do
     {
    
         m++;
         s1 = s;
         s = s - sqrt( m - 1) + sqrt(2 * m- 1) + sqrt (2*m);
     }
     while(s < n);
    printf(" 不等式的解为 m >= %ld\n",m);
    printf("注：当m = %ld时， s = %.2f; 当 m = %ld时， s = %.2f\n", m-1, s1, m, s);
    return 0;
}

结果：

//求解调和级数不等式 
 int main()
 {
     
    long c, d, m;
    double x, y, s;
    printf(" 请输入正数 x,y (2<x<y)： ");
    scanf ("%lf,%lf", &x, &y); 
       m=0;
	   s=0; 
	   while (s <= x) //循环求和探索m的下确界c 
         s = s + 1.0 / (++m);
       c = m;
        do
        s = s + 1.0/ (++m) ;
        while (s < y);//循环求和探索m的上确界d
        d = m-1;
        printf(" 满足不等式的解为：%1d ≤ m ≤ %1d \n", c, d);
        return 0;
 }

结果：

//解不等式：d<1+1/2-1/3+1/4+1/5-1/6+..士1/n 
int main()
{
     
    long n, k;
    double d,s;
     printf(" 请输入正数 d：");
     scanf ("%lf", &d) ;
      printf(" %lf <1+1/2-1/3+1/4+1/5-1/6+..+-1/n 的解：\n",d);
       n=3;
       s=3.0/2; 
       while (1)
        {
     
			s = s - 1.0 / n + 1.0 / (n+1) + 1.0 / (n+2);
        	if (s > d) break; 
          	n = n + 3;
        }
           printf(" n = %ld \n", n + 2);
           k = n + 2;
           while(1)
           {
    
                if ((s = s - 1.0 / (++k)) > d) break;
                 if ((s = s + 1.0 / (++k)) > d) break;
                  s = s + 1.0 / (++k);
                
                   printf(" n = %ld \n", k) ; //离散解k
           }
            printf(" n >= %ld \n", k);
            return 0;
}

结果：

 //解不等式：d<1+1/2-1/3+1/4+1/5-1/6+..士1/n 
int main()
{
     
    long n, k;
    double d,s;
     printf(" 请输入正数 d：");
     scanf ("%lf", &d) ;
      printf(" %lf <1+1/2-1/3+1/4+1/5-1/6+..+-1/n 的解：\n",d);
       n=1;
       s=0; 
       while (1)
        {
     
			s = s + 1.0 / n + 1.0 / (n+1) - 1.0 / (n+2);
        	if (s > d) break; 
          	n = n + 3;
        }
           printf(" n >= %ld \n", n + 1);
           for(s = 0, k =1; k <= n; k++)
           {
    
           	
           		if(k  % 3 > 0) s = s + 1.0 / k;
           		else s = s- 1.0/ k;
           		if(s > d)  printf(" n = %ld \n", k);
		   }
		   
           
            return 0;
}

结果：