高考问答lol鳄鱼怎么打盖伦:编程序.............
来源：百度文库 编辑：高考问答 时间：2024/04/29 14:01:58
-1 (x<0)
y= 0 (x=0)
1 (x>0)
2用for for 语句求出1+2+3+……100的和。
3画出2题中while结构的流程图
4求ax2+bx+c=0方程的根。A,b,c有键盘输入，设b2-4ac>0
把第三个做出来再给20分
1.#include <stdio.h>
int f(int n)
{
if (n>0) return 1;
if (n==0) return 0;
if (n<0) return -1;
}
main()
{
int a;
scanf("%d",&a);
printf("%d", f(a));
}
2.main( )
{ int i, sum=0;
for (i=1;i<=100;i++)
{ sum=sum+i; }
printf("sum=%d\n",sum); }
3.自己画吧
4.
#include"stdio.h"
#include"math.h"
void main()
{
float a,b,c,disc,x1,x2,p,q;
printf("input a,b,c:\n");
scanf("%f,%f,%f",&a,&b,&c);
if(a==0)
if(b==0)
if(c==0)
printf("It is trivial.\n");
else
printf("It is impossible.\n");
else
{
printf("It has one solution.\n");
printf("x=%6.2f\n",-c/b);
}
else
{
disc=b*b-4*a*c;
if(disc>=0)
if(disc>0)
{
printf("It has two real solutions:\n");
x1=(-b+sqrt(disc))/(2*a);
x2=(-b-sqrt(disc))/(2*a);
printf("x1=%6.2f,x2=%6.2f\n",x1,x2);
}
else
{
printf("It has two same real solutions:\n");
printf("x1=x2=%6.2f\n",-b/2*a);
}
else
{
printf("It has two complex solutions:\n");
p=-b/2*a;
q=(sqrt(-disc))/(2*a);
printf("x1=%6.2f+%6.2fi,x2=%6.2f-%6.2fi\n",p,q,p,q);
}
}
}