有了点状况，不是报错就是出现-1.#J，没查到错，特来请教

#include<stdio.h>
#include<math.h>
#include<string.h>
int solution(float a,float b,float c,float s[])  
{
    float d,x1,x2,sq;
    d = b*b - 4*a*c;
    sq = (float)sqrt(d);
    if(d >= 0)
    {
        if(d == 0)
            x1 = x2 = -b/(2*a);
        else
        {
            x1 = (-b+sq) / (2*a);
            x2 = (-b+sq) / (2*a);
        }
                s[0] = x1;
                s[1] = x2;
                return 1;
    }
    else
        return 0;
}
   
void main()
{
    float a,b,c;
    int i;
    float s[2];
    printf("请输入a,b,c:\n");
    scanf("%f,%f,%f",&a,&b,&c);
    if(solution(a,b,c,s))
        for(i = 0;i < 2;i++)
            printf("这个方程的解%d为：%.2f\n",i,s[i]);
    else
        printf("ERROR!!!\n");
}

        

            

#include <stdio.h>
#include <math.h>
#include <string.h>
int solution(float a,float b,float c,float s[])
{
    float d,sq;
    d=b*b-4*a*c;
    sq=sqrt(d);
    if(d>=0)
    {
        if(d == 0)
            s[0]=s[1]=-b/(2*a);
        else
        {
            s[0]=(-b+sq)/(2*a);
            s[1]=(-b-sq)/(2*a);
        }
        return 1;
    }
    else
        return 0;
}
int main()
{
    float a,b,c;
    int   i;
    float s[2];
    printf("请输入a,b,c:\n");
    scanf("%f%f%f",&a,&b,&c);
    if(solution(a,b,c,s))
        for(i=0; i<2; i++)
            printf("这个方程的解%d为：%.2f\n",i+1,s[i]);
    else
        printf("ERROR!!!\n");
    return 0;
}

现在我有一种蓦然回首那人却在灯火阑珊处的感觉，问题解决了

#include<stdio.h>
#include<math.h>
#include<string.h>
int solution(float a,float b,float c,float s[])   //solution(求解)
{
    float d,x1,x2,sq;
    d = b*b - 4*a*c;
    sq = (float)sqrt(d);
    if(d >= 0)
    {
        if(d == 0)
            x1 = x2 = -b/(2*a);
        else
        {
            x1 = (-b+sq) / (2*a);
            x2 = (-b-sq) / (2*a);
        }
                s[0] = x1;
                s[1] = x2;
                return 1;
    }
    else
        return 0;
}
   
void main()
{
    float a,b,c;
    int i;
    float s[2];
    printf("请输入a,b,c:\n");
    scanf("%f%f%f",&a,&b,&c);
    if(solution(a,b,c,s))
        for(i = 1;i < 3;i++)
            printf("这个方程的解%d为：%.2f\n",i,s[i - 1]);
    else
        printf("ERROR!!!\n");
}

        

            

比我快，呵呵