《算法竞赛入门经典》第四章小结

题目：编写一个函数solve，给定浮点数a,b,c,d,e,f，求解方程组ax+by=c, dx+ey=f.

任务1：使用assert宏，让解不唯一时异常退出。

任务2： 解不唯一时仍正常返回，但调用者有办法知道解的数量（无解，唯一解，无穷多组解）。

思考：函数的参数都有哪些，各是什么类型？

//任务1#include <stdio.h>#include <assert.h>double fucx(double a,double b,double c, double d,double e,double f){assert( (a*e-b*d)!=0 );return (c*e-b*f)/(a*e-b*d);} double fucy(double a,double b,double c, double d,double e,double f){assert( (a*e-b*d)!=0 );return (a*f-c*d)/(a*e-b*d);}int main(){double a,b,c,d,e,f;double x,y;scanf("%lf%lf%lf%lf%lf%lf",&a,&b,&c,&d,&e,&f);x=fucx(a,b,c,d,e,f);printf("\n");y=fucy(a,b,c,d,e,f);printf("%lf %lf\n",x,y);return 0;}

//任务2#include <stdio.h>#include <assert.h>int decide(double a,double b,double c, double d,double e,double f){if(a*e==b*d){if( (c*e==b*f)&&(a*f==c*d) ){return 2;}else{return 3;}}elsereturn 1;} int main(){double a,b,c,d,e,f;double x,y;scanf("%lf%lf%lf%lf%lf%lf",&a,&b,&c,&d,&e,&f);int out;out=decide(a,b,c,d,e,f);if(out==1){printf("%lf %lf\n",(c*e-b*f)/(a*e-b*d),(a*f-c*d)/(a*e-b*d));}if(out==2)printf("There is infinite solution.\n");if(out==3)printf("There is no solution.\n");return 0;}