算法提高 复数归一化
时间限制:1.0s 内存限制:512.0MB
编写函数Normalize,将复数归一化,即若复数为a+bi,归一化结果为a/sqrt(a*a+b*b) + i*b/sqrt(a*a+b*b) 。使用结构体指针类型作为函数参数可能是必要的。其中实部和虚部由键盘输入,输出为归一化结果,如果归一化结果的实部或虚部为小数的要求保留一位小数。
样例输入:(格式说明:3 4 分别为以空格隔开的实数的实部和虚部)
3 4
样例输出:
0.6+0.8i
样例输入:
2 5
样例输出:
0.4+0.9i
#include <stdio.h>
#include <math.h>
struct ComplexNumber
{
double real;
double img;
};
void Normalize(struct ComplexNumber *num)
{
double norm_real = num->real / sqrt(num->real * num->real + num->img * num->img);
double norm_img = num->img / sqrt(num->real * num->real + num->img * num->img);
num->real = norm_real;
num->img = norm_img;
}
int main()
{
struct ComplexNumber number;
scanf("%lf %lf", &number.real, &number.img);
Normalize(&number);
if (number.real == (int)number.real)
printf("%d", (int)number.real);
else
printf("%.1lf", number.real);
if (number.img == (int)number.img)
printf("%+di", (int)number.img);
else
printf("%+.1lfi", number.img);
return 0;
}