算法训练 P1103

#include <cstdio>
#include <iostream>

using namespace std;

struct num{
    double a;
    double b;
};
void add(struct num A, struct num B, struct num* C);
void sub(struct num A, struct num B, struct num* C);
void mul(struct num A, struct num B, struct num* C);
void div(struct num A, struct num B, struct num* C);
void clac(char op, struct num A, struct num B, struct num* C);

void clac(char op, struct num A, struct num B, struct num* C){

    switch(op){
        case '+': add(A,B,C); break;
        case '-': sub(A,B,C); break;
        case '*': mul(A,B,C); break;
        case '/': div(A,B,C); break;
    }
}

void add(struct num A, struct num B, struct num* C){
    C->a = A.a + B.a;
    C->b = A.b + B.b;
}

void sub(struct num A, struct num B, struct num* C){
    C->a = A.a - B.a;
    C->b = A.b - B.b;
}

void mul(struct num A, struct num B, struct num* C){
    C->a = A.a*B.a - A.b*B.b;
    C->b = A.a*B.b + A.b*B.a;
}

void div(struct num A, struct num B, struct num* C){
    double x = B.a*B.a + B.b*B.b;
    C->a = (A.a*B.a + A.b*B.b)/x;
    C->b = (A.b*B.a - A.a*B.b)/x;
}

int main(){
    struct num A;
    struct num B;
    struct num C;
    char op;
    cin >> op >> A.a >> A.b >> B.a >> B.b;
    clac(op, A,B,&C);
    printf("%.2f+%.2fi
", C.a, C.b);

    return 0;
}