【PAT】A1055 The World‘s Richest

最新推荐文章于 2021-09-24 14:46:09 发布

原创最新推荐文章于 2021-09-24 14:46:09 发布 · 192 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#PTA

—— AC笔记 —— 专栏收录该内容

67 篇文章

订阅专栏

本博客介绍了一种算法挑战，即从全球富豪榜中筛选出特定年龄段内的最富有人群。通过解析输入数据，使用结构体存储姓名、年龄和财富信息，并实现定制的排序算法进行高效筛选。

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 Npeople, 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 (≤105) - the total number of people, and K (≤103) - 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 [−106,106]) 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

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

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

解题思路：

这是一道排序中的常规题，读完题目后的解题程序和思考方式为：

创建结构体，包括名字数组（比题目给的8个空间多一位），年龄和财富参数
按题意创建比较数组cmp()，也比较常规。但是由于要比较字符数组，建议先用strcmp()函数赋值给一个变量再去判断变量与0的关系，会好很多，也是一个习惯；
输入参数，比较；
最后选择排序中，运用条件判断和计数器count，当count==num时（count初值为0），而则跳出当前轮查询，边查询边输出有利于不用再去定义新的空间来存放已有变量。
检验计数器count的值，当count=0时，也就意味着找不到区间中的人，所以输出None，进入下一轮查询。

代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

struct per{
	char name[9];
	int age,weath;
	bool flat;
}p[100010];

bool cmp(per a,per b){
	int s=strcmp(a.name,b.name);
	if(a.weath!=b.weath) return a.weath>b.weath;
	else if(a.age!=b.age) return a.age<b.age;
	else if(s!=0) return s<0;
}
int main(){
	int n,k;
	scanf("%d%d",&n,&k);
	for(int i=0;i<n;i++){
		scanf("%s%d%d",p[i].name,&p[i].age,&p[i].weath);
	}
	sort(p,p+n,cmp);
	int num,Amin,Amax; 
	
	for(int i=0;i<k;i++){//查询轮数 
		scanf("%d%d%d",&num,&Amin,&Amax);
		int count=0;// 计数器每一轮清零 
		printf("Case #%d:\n",i+1);
		for(int j=0;j<n&&count<num;j++){//每一轮查找 
			if(p[j].age>=Amin&&p[j].age<=Amax){
				printf("%s %d %d\n",p[j].name,p[j].age,p[j].weaath);
				count++;
			}
		}
		if(count==0) printf("None\n"); 
	}
	return 0;
}