// equation.cpp : Defines the entry point for the console application.//#include <stdio.h>#include <stdlib.h>#include <math.h>#include <float.h> // 定义了浮点型的无穷小常量 const float FLT_EPSILON=1.192092896e-07F;/** The function to caculate a Real coefficient eqution’s(ax^2+bx+c=0) root.* IN: a,b,c —- the three real coefficient,* OUT: r1, r2 —- the two real roots or the real part of the complex roots.* i1, i2 —- the image part of the complex roots.* RET: the status of the equation.*/int equation1(float a, float b, float c, float* r1, float* r2, float* i1, float* i2);int equation2(float a, float b, float c, float& r1, float& r2, float& i1, float& i2);int main(int argc, char* argv[]){ float a; float b; float c; char xuanze; float r1; float r2; float i1; float i2; for(;;) { printf(“是否进行一元二次方程计算？Y/N\n”); scanf(“%c”,&xuanze); if(xuanze==’Y’) { printf(“Please input a,b,c of the equation ax^2+bx+c=0 [a b c Enter]: “); scanf(“%f %f %f”, &a, &b, &c); switch(equation2(a,b,c, r1, r2, i1, i2)) { case 0: printf(“The equation %gx^2+%gx+%g=0 has no roots.\n”, a, b, c); break; case 1: printf(“The equation %gx^2+%gx+%g=0 has one real root:\n \tx=%g.\n”, a, b, c, r1); break; case 2: printf(“The equation %gx^2+%gx+%g=0 has two real roots:\n \tx1=%g, \tx2=%g.\n”, a, b, c, r1, r2); break; case 3: printf(“The equation %gx^2+%gx+%g=0 has an arbitrary solution.\n”, a, b, c); break; case 4: printf(“The equation %gx^2+%gx+%g=0 has a pair of conjugate complex roots:\n\tx1=%g+%gi, \tx2=%g-%gi.\n”, a, b, c, r1, i1, r1, i1); break; default: break; } _flushall(); } else { return 0; } } }// printf(“选择的是%c\n”,xuanze);/** The function to caculate a Real coefficient eqution’s(ax^2+bx+c=0) root.* IN: a,b,c —- the three real coefficient,* OUT: r1, r2 —- the two real roots or the real part of the complex roots.* i1, i2 —- the image part of the complex roots.* RET: the status of the equation.* 0 —- no solution.* 1 —- one real root.* 2 —- two real root.* 3 —- has an arbitrary solution.* 4 —- has a pair of conjugate complex roots.*/int eqution1(float a, float b, float c, float* r1, float* r2, float* image){ return 0;}/** The function to caculate a Real coefficient eqution’s(ax^2+bx+c=0) root.* IN: a,b,c —- the three real coefficient,* OUT: r1, r2 —- the two real roots or the real part of the complex roots.* i1, i2 —- the image part of the complex roots.* RET: the status of the equation.* 0 —- no solution.* 1 —- one real root.* 2 —- two real root.* 3 —- has an arbitrary solution.* 4 —- has a pair of conjugate complex roots.*/int equation2(float a, float b, float c, float& r1, float& r2, float& i1, float& i2){ int s; double delta; if (0==a){ if(0==b){ if(0==c){ s= 3; } else{ s=0; } } else{ r1=-c/b; s=1; } } else { delta=b*b-4.0f*a*c; if(delta>=-FLT_EPSILON && delta<=FLT_EPSILON) // delta==0 { r1=r2=-b/2.0f/a; s=1; } else if(delta>FLT_EPSILON){ // delta>0 r1=-b/2.0f/a+(float)sqrt(delta); r2=-b/2.0f/a-(float)sqrt(delta); s=2; } else{ // delta<0 r1=r2=-b/2.0f/a; i1=(float)sqrt(-delta); i2=-(float)sqrt(-delta); s=4; } } return s;}
———————————————————————————————————————————————————————————————————————————这个程序用来实现用户判断是否进行一元二次方程计算的循环，程序结构简单，写的过程中遇到问题在scanf方法，第一次循环时scanf方法正确获取用户输入，但第二次循环的时候获取错误。了解了scanf（）方法后，发现scanf每次接收的内容是 输入+回车键。所以可以使用flushll（）；清空缓存区。经历一种身体下了地狱，眼睛进入天堂，灵魂归入故里。
