public double (doubloe a ,double b ){ return a + b; }
public double (doubloe s ,double b ){ return s + b; }
public double sum(double a,double b) { return a+b; }
public static double add(double v1,double v2) { BigDecimal b1 = new BigDecimal(Double.toString(v1)); BigDecimal b2 = new BigDecimal(Double.toString(v2)); return b1.add(b2).doubleValue(); }
public static String add(String a, String b) { return new BigInteger(a).add(new BigInteger(b)).toString(); }or public static String add(String a, String b) { return new BigDecimal(a).add(new BigDecimal(b)).toString(); }
zcjl 的方法就可以了啊—— BigInteger不限制位数的
呵呵,忘了说了,不能用BigInteger的
public static String add(String s, String t) { int a=0,b=0; int s0 = s.length(); int t0 = t.length(); int i,m; int n=Math.max(s0,t0) + 1; char[] result = new char[n]; for (i=n-1; i>=0; i--) { if (i+s0-n>=0) a=Integer.parseInt(s.substring(i+s0-n,i+s0-n+1)); else a=0; if (i+t0-n>=0) b=Integer.parseInt(t.substring(i+t0-n,i+t0-n+1)); else b=0; m = a + b; if (m>=10) { m-=10; result[i-1] += 1; } result[i] += '0' + m; } return new String(result); }
java时面又没有模板,怎么实现啊,你以为是C++啊 可以用函数重载解决这个问题吗
写一个java类 在类中有+,-,*,/方法,大致是这样,实部与虚部相加(具体复数是否这样可以相加,本人有点忘记,但设计思想应该是这样.) class Calcute { private double a; //实部 private double b; //虚部 public Calcute (double a,double b) { this.a =a ; this.b = b; } public Calcute add(Calcute obj) { this.a + obj.a; this.b + obj.b; return new Calcute(a,b); } ... } public class Test { public static void main(String[] args) { Calcute c1 = new Calcute(1,2); Calcute c2 = new Calcute(3,4); Calcute c3 = c1.add(c2); } }
public double sum(double a,double b) { return a+b; }
return a + b;
}
return s + b;
}
{
return a+b;
}
{
BigDecimal b1 = new BigDecimal(Double.toString(v1));
BigDecimal b2 = new BigDecimal(Double.toString(v2));
return b1.add(b2).doubleValue();
}
return new BigInteger(a).add(new BigInteger(b)).toString();
}or public static String add(String a, String b) {
return new BigDecimal(a).add(new BigDecimal(b)).toString();
}
BigInteger不限制位数的
int a=0,b=0;
int s0 = s.length();
int t0 = t.length();
int i,m;
int n=Math.max(s0,t0) + 1;
char[] result = new char[n];
for (i=n-1; i>=0; i--)
{ if (i+s0-n>=0) a=Integer.parseInt(s.substring(i+s0-n,i+s0-n+1)); else a=0;
if (i+t0-n>=0) b=Integer.parseInt(t.substring(i+t0-n,i+t0-n+1)); else b=0; m = a + b;
if (m>=10)
{
m-=10;
result[i-1] += 1;
}
result[i] += '0' + m;
}
return new String(result);
}
可以用函数重载解决这个问题吗
在类中有+,-,*,/方法,大致是这样,实部与虚部相加(具体复数是否这样可以相加,本人有点忘记,但设计思想应该是这样.)
class Calcute {
private double a; //实部
private double b; //虚部
public Calcute (double a,double b) {
this.a =a ;
this.b = b;
} public Calcute add(Calcute obj) {
this.a + obj.a;
this.b + obj.b;
return new Calcute(a,b);
}
...
}
public class Test {
public static void main(String[] args) {
Calcute c1 = new Calcute(1,2);
Calcute c2 = new Calcute(3,4);
Calcute c3 = c1.add(c2);
}
}
{
return a+b;
}