1055 The World's Richest（25 分）

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ChenyutingZJU

最新推荐文章于 2025-03-11 11:28:52 发布

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博客围绕模拟福布斯富豪排名查询问题展开，给出输入输出规格及示例。解题思路为用结构体数组记录信息并设置排序函数，读入数据后排序，根据人数和年龄段限制遍历数组输出结果，若未找到符合条件的人则输出None。

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1055 The World’s Richest（25 分）
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
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：
1.用结构体数组记录各项信息，设置cmp排序函数。
2.按要求读入，并排序。
3.根据输出对人数和年龄段的限制，遍历排序后的数组，符合要求就输出，达到制定人数就退出。
4.每一个查询，设置cnt记录已经输出的人数，如果为0，要额外输出None

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1e5+10;
struct Node{
    int age,worth;
    char name[10];
}node[maxn];

bool cmp(Node a,Node b){
    if(a.worth !=b.worth )
        return a.worth >b.worth ;
    else if(a.age !=b.age )
        return a.age <b.age ;
    else 
        return strcmp(a.name ,b.name )<0;
}

int main(){
    int n,m,k,amin,amax;
    scanf("%d%d",&n,&k);
    for(int i=0;i<n;i++){
        scanf("%s %d %d",node[i].name ,&node[i].age ,&node[i].worth );
    }
    sort(node,node+n,cmp);
    for(int i=1;i<=k;i++){
        scanf("%d%d%d",&m,&amin,&amax);
        printf("Case #%d:\n",i);
        int cnt=0;
        for(int j=0;j<n;j++){
            if(node[j].age >=amin&&node[j].age <=amax){
                printf("%s %d %d\n",node[j].name,node[j].age ,node[j].worth );
                cnt++;
            }
            if(cnt==m) break;
        }
        if(cnt==0) printf("None\n");
    }
    return 0;
}