1055. The World’s Richest (25)
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 (<=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 50Sample 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
分析：这道题目，是对数据进行排序，然后按照要求输出。挺简单的。
code:
/*************************************************************************> Created Time: 2017年09月12日 星期二 19时47分25秒************************************************************************/
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<vector>
#include<cstring>
#include<algorithm>
struct man{char name[10];int age;long wealth;
};
bool comp(man a,man b)
{if(a.wealth>b.wealth)return true;else if(a.wealth<b.wealth)return false;if(a.age<b.age)return true;if(a.age>b.age)return false;if(strcmp(a.name,b.name)<0)return true;return false;
}
using namespace std;
vector<man>men;
int main()
{freopen("in","r",stdin);int N,K;man tmp;scanf("%d%d",&N,&K);for(int i=0;i<N;i++){scanf("%s%d%ld",tmp.name,&tmp.age,&tmp.wealth);men.push_back(tmp);}sort(men.begin(),men.end(),comp);int c,a,b,count;for(int i=0;i<K;i++){count=0;scanf("%d%d%d",&c,&a,&b);printf("Case #%d:\n",i+1);for(int j=0;j<men.size();j++){if(men[j].age>=a&&men[j].age<=b){count++;printf("%s %d %ld\n",men[j].name,men[j].age,men[j].wealth);if(count==c)break;}}if(count==0)printf("None\n");}return 0;
}
- AC:
