题目描述

求方程 的根，用三个函数分别求当b^2-4ac大于0、等于0、和小于0时的根，并输出结果。从主函数输入a、b、c的值。

输入

a b c

输出

x1=? x2=?

样例输入

4 1 1

样例输出

x1=-0.125+0.484i x2=-0.125-0.484i

1 #include 
     
       2 #include 
      
        3  4 // delta = 0 5 void delta_eq_0(int a, int b, int c) 6 { 7     float res = (float)(0 - b) / (2 * a); 8     printf("x1=%.3f x2=%.3f\n", res, res); 9 }10 11 // delta > 012 void delta_gt_0(int a, int b, int c)13 {14     int delta = b * b - 4 * a * c;15     float res1, res2;16     res1 = (-b+sqrt(delta)) / (2 * a);17     res2 = (-b-sqrt(delta)) / (2 * a);18     printf("x1=%.3f x2=%.3f\n", res1, res2);19 }20 21 // delta < 022 void delta_lt_0(int a, int b, int c)23 {24     int delta = b * b - 4 * a * c;25     float res1, res2;26     res1 = (float)(0 - b) / (2 * a);27     res2 = sqrt(abs(delta)) / (2 * a);28     printf("x1=%.3f+%.3fi x2=%.3f-%.3fi\n", res1, res2, res1, res2);29 }30 31 // 计算结果32 void caculate(int a, int b, int c)33 {34     int delta = b * b - 4 * a * c;35     36     if(delta == 0)    37         delta_eq_0(a, b, c);38     else if(delta > 0)39         delta_gt_0(a, b, c);40     else        41         delta_lt_0(a, b, c);42 }43 44 45 int main(int argc, char const *argv[])46 {47 48     int a, b, c;49     scanf("%d%d%d", &a, &b, &c);50     caculate(a, b, c);51     return 0;52 }