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PAT甲级-1055 The World's Richest (25分)

  • 24-03-03 00:44
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blog.csdn.net

点击链接PAT甲级-AC全解汇总

题目:
Forbes magazine publishes every year its list of billionaires based on the annual ranking of the world’s wealthiest people. Now you are supposed to simulate this job, but concentrate only on the people in a certain range of ages. That is, given the net worths of N people, you must find the M richest people in a given range of their ages.

Input Specification:
Each input file contains one test case. For each case, the first line contains 2 positive integers: N (≤10​5​​ ) - the total number of people, and K (≤10​3​​ ) - the number of queries. Then N lines follow, each contains the name (string of no more than 8 characters without space), age (integer in (0, 200]), and the net worth (integer in [−10​6​​ ,10​6​​ ]) of a person. Finally there are K lines of queries, each contains three positive integers: M (≤100) - the maximum number of outputs, and [Amin, Amax] which are the range of ages. All the numbers in a line are separated by a space.

Output Specification:
For each query, first print in a line Case #X: where X is the query number starting from 1. Then output the M richest people with their ages in the range [Amin, Amax]. Each person’s information occupies a line, in the format

Name Age Net_Worth
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The outputs must be in non-increasing order of the net worths. In case there are equal worths, it must be in non-decreasing order of the ages. If both worths and ages are the same, then the output must be in non-decreasing alphabetical order of the names. It is guaranteed that there is no two persons share all the same of the three pieces of information. In case no one is found, output None.

Sample Input:

12 4
Zoe_Bill 35 2333
Bob_Volk 24 5888
Anny_Cin 95 999999
Williams 30 -22
Cindy 76 76000
Alice 18 88888
Joe_Mike 32 3222
Michael 5 300000
Rosemary 40 5888
Dobby 24 5888
Billy 24 5888
Nobody 5 0
4 15 45
4 30 35
4 5 95
1 45 50
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Sample Output:

Case #1:
Alice 18 88888
Billy 24 5888
Bob_Volk 24 5888
Dobby 24 5888
Case #2:
Joe_Mike 32 3222
Zoe_Bill 35 2333
Williams 30 -22
Case #3:
Anny_Cin 95 999999
Michael 5 300000
Alice 18 88888
Cindy 76 76000
Case #4:
None
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题意:
输入一堆人的年龄和身价
对于每个查询,输出查询范围内前n个人的信息
按照财富大,年龄小,人名小的顺序输出,没有的话None

注意: 可能会超时,一样的代码,我提交了十多次,有两次可以过
已经不用cout了,也把class改成了struct。。。。

反正没有罚时~~

在这里插入图片描述
在这里插入图片描述

我的代码:

#include
using namespace std;

struct Person{
    string name;
    int age;
    int worth;
};
bool mycmp(Person& a,Person& b){
    if(a.worth!=b.worth)return a.worth>b.worth;
    else if(a.age!=b.age)return a.age<b.age;
    else return a.name<b.name;
}
void myprint(Person& a){
    printf("%s %d %d\n",a.name.c_str(),a.age,a.worth);
}

int main()
{
    int N,K;
    scanf("%d %d",&N,&K);
    vector<Person>alls;
    for(int i=0;i<N;i++)
    {
        string name;
        int age,worth;
        cin>>name>>age>>worth;
        alls.push_back(Person{name,age,worth});
    }
    sort(alls.begin(),alls.end(),mycmp);
    for(int i=1;i<=K;i++)
    {
        int n,min_age,max_age,index=0,cnt=0;
        scanf("%d %d %d",&n,&min_age,&max_age);
        printf("Case #%d:\n",i);
        while(cnt<n&&index<N)
        {
            int age=alls[index].age;
            if(age>=min_age&&age<=max_age)
            {
                myprint(alls[index]);
                cnt++;
            }
            index++;
        }
        if(!cnt)printf("None\n");
    }

    return 0;
}

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注:本文转载自blog.csdn.net的邂逅模拟卷的文章"https://blog.csdn.net/qq_34451909/article/details/105656902"。版权归原作者所有,此博客不拥有其著作权,亦不承担相应法律责任。如有侵权,请联系我们删除。
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