复数的简单运算
2013-05-25 17:55
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模板: 方法名(A的实部,A的虚部,B的实部,B的虚部,结果的实部,结果的虚部)
公式:乘法 (a*c - b* d) + (b * c + a * d)i
除法 (a* c + b* d )/(c*c + d* d) + (b * c - a * d )/(c * c + d* d)i
代码如下:
package com.njupt.acm;
public class Test2 {
public static void main(String[] args) {
double a = 4 , b = 6 , c = 2 , d = -1;
double[] e = {0} , f = {0};
cAdd(a,b,c,d , e , f);
show(a, b, c, d, e, f);
cSub(a,b,c,d,e,f);
show(a, b, c, d, e, f);
cMul(a,b,c,d,e,f);
show(a, b, c, d, e, f);
cDiv(a,b,c,d,e,f);
show(a, b, c, d, e, f);
}
public static void cAdd(double a, double b, double c, double d, double[] e,
double[] f) {
e[0] = a + c;
f[0] = b + d;
}
public static void cSub(double a, double b, double c, double d, double[] e,
double[] f) {
e[0] = a - c;
f[0] = b - d;
}
public static void cMul(double a, double b, double c, double d, double[] e,
double[] f) {
e[0] = a * c - b * d;
f[0] = b * c + a * d;
}
public static void cDiv(double a, double b, double c, double d, double[] e,
double[] f) {
double sq = c * c + d * d;
e[0] = (a * c + b * d) / sq;
f[0] = (b * c - a * d) / sq;
}
public static void show( double a , double b , double c , double d , double[] e , double[] f){
System.out.println("("+a +" + "+ b + "i) + " + "(" + c + " + "+ d + "i) = " + e[0] +" + "+ f[0] + "i" );
}
}
公式:乘法 (a*c - b* d) + (b * c + a * d)i
除法 (a* c + b* d )/(c*c + d* d) + (b * c - a * d )/(c * c + d* d)i
代码如下:
package com.njupt.acm;
public class Test2 {
public static void main(String[] args) {
double a = 4 , b = 6 , c = 2 , d = -1;
double[] e = {0} , f = {0};
cAdd(a,b,c,d , e , f);
show(a, b, c, d, e, f);
cSub(a,b,c,d,e,f);
show(a, b, c, d, e, f);
cMul(a,b,c,d,e,f);
show(a, b, c, d, e, f);
cDiv(a,b,c,d,e,f);
show(a, b, c, d, e, f);
}
public static void cAdd(double a, double b, double c, double d, double[] e,
double[] f) {
e[0] = a + c;
f[0] = b + d;
}
public static void cSub(double a, double b, double c, double d, double[] e,
double[] f) {
e[0] = a - c;
f[0] = b - d;
}
public static void cMul(double a, double b, double c, double d, double[] e,
double[] f) {
e[0] = a * c - b * d;
f[0] = b * c + a * d;
}
public static void cDiv(double a, double b, double c, double d, double[] e,
double[] f) {
double sq = c * c + d * d;
e[0] = (a * c + b * d) / sq;
f[0] = (b * c - a * d) / sq;
}
public static void show( double a , double b , double c , double d , double[] e , double[] f){
System.out.println("("+a +" + "+ b + "i) + " + "(" + c + " + "+ d + "i) = " + e[0] +" + "+ f[0] + "i" );
}
}
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