（改）~-4 复数四则运算

以下是我的错误代码（失了智）

#include<cstdio>
using namespace std;

double q,w;
void p(double a, double b, double c, double d){
    double jj,kk;
    double j = a + c;
    double k = b + d;
    //printf("%f    %f \n", j ,k);
    if(j < 0) jj = -j;
    if(k < 0) kk = -k;
    if( jj<0.05  && kk<0.05)
    printf("(%.1f+%.1fi) + (%.1f-%.1fi) = %.1f\n", a, b, c, d, 0.0);
    else if(jj<0.05) printf("(%.1f+%.1fi) + (%.1f-%.1fi) = %.1fi\n", a,b,c,d,k);
    else if(kk<0.05) printf("(%.1f+%.1fi) + (%.1f-%.1fi) = %.1f\n", a,b,c,d,j);
    else
        printf("(%.1f+%.1fi) + (%.1f-%.1fi) = %.1f+%.1fi\n", a,b,c,d,j,k);
}

void de(double a, double b, double c, double d){
    double jj,kk;
    double j = a - c;
    double k = b - d;
    if(j < 0) jj = -j;
    if(k < 0) kk = -k;
    if(jj<0.05&& kk<0.05)
    printf("(%.1f+%.1fi) - (%.1f-%.1fi) = %.1f\n", a, b, c, d, 0.0);
    else if(jj<0.05) printf("(%.1f+%.1fi) - (%.1f-%.1fi) = %.1fi\n", a,b,c,d,k);
    else if(kk<0.05) printf("(%.1f+%.1fi) - (%.1f-%.1fi) = %.1f\n", a,b,c,d,j);
    else
        printf("(%.1f+%.1fi) - (%.1f-%.1fi) = %.1f+%.1fi\n", a,b,c,d,j,k);
}

void x(double a, double b, double c, double d){
    q = a * c - b * d;
    w = a * d + b * c;
}

void chu(double a, double b, double c, double d){
    double qq,ww;
    double j = c * c + d * d;
    x(a, b, c , -d);
    q = q/j;
    w = w/j;
    if(q < 0) qq = -q;
    if(w < 0) ww = -w;
    if(q<0.05&& w<0.05)
    printf("(%.1f+%.1fi) / (%.1f-%.1fi) = %.1f\n", a, b, c, d, 0.0);
    else if(qq<0.05) printf("(%.1f+%.1fi) / (%.1f-%.1fi) = %.1fi\n", a,b,c,d,w);
    else if(ww<0.05) printf("(%.1f+%.1fi) / (%.1f-%.1fi) = %.1f\n", a,b,c,d,q);
    else
        printf("(%.1f+%.1fi) / (%.1f-%.1fi) = %.1f+%.1fi\n", a,b,c,d,q,w);
}

int main(){
    double a,b,c,d;
    while( scanf("%lf %lf %lf %lf", &a, &b ,&c ,&d) == 4) {
        double qq,ww;
        p(a,b,c,d);
        de(a,b,c,d);
        x(a,b,c,d);
        if(q < 0) qq = -q;
        if(w < 0) ww = -w;
        if(qq<0.05&& ww<0.05)
            printf("(%.1f+%.1fi) * (%.1f-%.1fi) = %.1f\n", a, b, c, d, 0.0);
        else if(qq<0.05) printf("(%.1f+%.1fi) * (%.1f-%.1fi) = %.1fi\n", a,b,c,d,w);
        else if(ww<0.05) printf("(%.1f+%.1fi) * (%.1f-%.1fi) = %.1f\n", a,b,c,d,q);
        else
            printf("(%.1f+%.1fi) * (%.1f-%.1fi) = %.1f+%.1fi\n", a,b,c,d,q,w);
        chu(a,b,c,d);
    }
    return 0;
}

以下是学长的代码：

#include <cstdio>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>
#include <string>
#include <map>
#include <set>
#include <queue>

using namespace std;
#define For(i,l,r) for(int (i)=(int)(l);(i)<(int)(r);++(i))
#define FOR(i,l,r) for(int (i)=(int)(l);(i)<=(int)(r);++(i))
#define rof(i,l,r) for(int (i)=(int)(l);(i)>=(int)(r);--(i))
typedef pair<int,int> pii;
typedef long long ll;
typedef unsigned long long llu;
typedef unsigned int ui;

#define eps 0.05
double a1,b1,a2,b2;
void out(double a,double b){
   // cout<<a<<' '<<b<<endl;
    if(fabs(a)<eps&&fabs(b)<eps)printf("0.0");
    else if(fabs(a)<eps)printf("%.1fi",b);
    else if(fabs(b)<eps)printf("%.1f",a);
    else{
        printf("%.1f",a);
        if(b>eps)printf("+");
        printf("%.1fi",b);
    }
}
void print(double a,double b){
    printf("(");out(a,b);printf(")");
}
void solve(){
    print(a1,b1);printf(" + ");print(a2,b2);printf(" = ");out(a1+a2,b1+b2);puts("");
    print(a1,b1);printf(" - ");print(a2,b2);printf(" = ");out(a1-a2,b1-b2);puts("");
    print(a1,b1);printf(" * ");print(a2,b2);printf(" = ");out(a1*a2-b1*b2,a1*b2+a2*b1);puts("");
    print(a1,b1);printf(" / ");print(a2,b2);printf(" = ");
    double fm=a2*a2+b2*b2;
    out((a1*a2+b1*b2)/fm,(b1*a2-a1*b2)/fm);puts("");
}
int main(){
    //freopen("in.txt","r",stdin);
    scanf("%lf%lf%lf%lf",&a1,&b1,&a2,&b2);
    solve();
    return 0;
}

其实我的思路问题不大，就是代码写得少了，经验不足，这里有重复性的步骤，我应该要想到用函数的，然后fabs（）我却用了复杂的转化。。。最后是我的代码没有考虑b为－的时候表示的问题。