题解 | #复数集合#

//输出最大的可想到大根堆，所以采用优先队列，又是复数，重载'<'(小于号)即可。
#include "stdio.h"
#include "queue"
#include "string"
using namespace std;
struct plural{
    int real;
    int fake;
};

bool operator <(plural lhs,plural rhs){//小于号重载
    int module1 = lhs.real*lhs.real + lhs.fake*lhs.fake;
    int module2 = rhs.real*rhs.real + rhs.fake*rhs.fake;
    if (module1 < module2)
        return true;
    else if (module1 == module2){
        if (lhs.fake > rhs.fake)
            return true;
        else
            return false;
    } else{
        return false;
    }
}

void myPQueuePop(priority_queue<plural> &myPQueue){//复数的弹出
    if (myPQueue.empty()){
        printf("empty\n");
    } else{
        plural e = myPQueue.top();
        printf("%d+i%d\n",e.real,e.fake);
        myPQueue.pop();
        printf("SIZE = %d\n",myPQueue.size());
    }
}

void insertPlural(priority_queue<plural> &myPQueue,plural e){//复数的插入
    myPQueue.push(e);
    printf("SIZE = %d\n",myPQueue.size());
}

plural findE(string str){//识别出复数
    int sum1 = 0;int sum2 = 0;//sum1为real，sum2为fake
    int i = 7;
    while (str[i] >= '0' && str[i] <= '9'){
        sum1 = sum1*10+str[i]-'0';
        ++i;
    }
    i = str.find('i');
    ++i;
    while (str[i] >= '0' && str[i] <= '9'){
        sum2 = sum2*10+str[i]-'0';
        ++i;
    }
    plural e;
    e.real = sum1;e.fake = sum2;
    return e;
}

int main(){
    int n;char buf1[100];
    priority_queue<plural> myPQueue;
    scanf("%d\n",&n);
    for (int i = 0; i < n; ++i) {
        fgets(buf1,30,stdin);
        string str = buf1;str.pop_back();
        if (str == "Pop"){
            myPQueuePop(myPQueue);
        } else{
            plural e = findE(str);
            insertPlural(myPQueue,e);
        }
        str.clear();
    }
}