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 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

生詞

題目大意：

給出n個人的姓名，年齡和擁有的錢，讓步進行k次查詢，每次查詢輸出在年齡區間內的財富值的從大到小的前m個人的資訊。如果財富值相同就就先輸出年齡小的，如果年齡相同就把名字按照字典序排序輸出

分析：

不能先排序然後根據沒一個條件再新建一個數組，對新陣列排序的方法，這樣測試點2會超時～因為n和m的懸殊太大了，n有10的5次方，m卻只有100個。所以先把所有的人按照財富值排序，再建立一個數組book標記每個年齡段擁有的人的數量，遍歷陣列並統計相應年齡的人數，噹噹前年齡的人的數量不超過100的時候壓入新的陣列，否則就不要壓入新的陣列。因為無論怎樣的年齡區間，查詢人數都不會超過100個人，所以最多隻要取每個年齡的前100個財富值的人到新的數組裡面就可，再從這個新的數組裡面取符合相應年齡的人的資訊～

原文連結：https://blog.csdn.net/liuchuo/article/details/52225204

測試點2超時程式碼

#include <bits/stdc++.h>

using namespace std;
struct person
{
    string name;
    int age,worth;
};
bool cmp(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;
    }
}
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif
    int n,k;
    scanf("%d%d",&n,&k);
    vector<person> v(n);
    for(int i=0; i<n; i++)
    {
        //scanf("%s %d %d",v[i].name.c_str(),&v[i].age,&v[i].worth);
        //printf("%s\n",v[i].name.c_str());
        cin>>v[i].name>>v[i].age>>v[i].worth;
    }
    sort(v.begin(),v.end(),cmp);
    for(int i=1; i<=k; i++)
    {
        int num,a,b,cnt=0;
        scanf("%d %d %d",&num,&a,&b);
        printf("Case #%d:\n",i);
        for(int j=0; j<n; j++)
        {
            if(v[j].age>=a&&v[j].age<=b)
            {
                if(cnt>=num)
                    break;
                cnt++;
                //printf("%s %d %d\n",v[j].name.c_str(),v[j].age,v[j].worth);
                cout<<v[j].name<<" "<<v[j].age<<" "<<v[j].worth<<endl;
            }
        }
        if(!cnt) printf("None\n");
    }
    return 0;
}

本來以為是cin、cout的事，後來發現不是。

新的困惑

關於C++中string的scanf與printf的問題：

#include <stdio.h>
#include <string>
using namespace std;
int main()
{
	string a;
	a.resize(100); //需要預先分配空間
	scanf("%s", &a[0]);
	puts(a.c_str());
	return 0;
}

一般應該是隻能printf可以使用a.c_str()。
但是我測試的上述程式碼的註釋的兩行，用scanf使用a.c_str()，直接輸出是有值的，但是後面的就輸不出了，，，

scanf("%s %d %d",v[i].name.c_str(),&v[i].age,&v[i].worth); printf("%s\n",v[i].name.c_str());

Zoe_Bill
Bob_Volk
Anny_Cin
Williams
Cindy
Alice
Joe_Mike
Michael
Rosemary
Dobby
Billy
Nobody

題解

柳神方法的別的博主的解釋：
原文連結：https://blog.csdn.net/qq_42437577/article/details/104133667

#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>
using namespace std;
struct node {
    char name[10];
    int age, money;
};
int cmp1(node a, node b) {
    if(a.money != b.money)
        return a.money > b.money;
    else if(a.age != b.age)
        return a.age < b.age;
    else
        return (strcmp(a.name, b.name) < 0);
}
 
int main() {
    int n, k, num, amin, amax;
    scanf("%d %d", &n, &k);
    vector<node> vt(n), v;
    vector<int> book(205, 0);
    for(int i = 0; i < n; i++)
        scanf("%s %d %d", vt[i].name, &vt[i].age, &vt[i].money);
    sort(vt.begin(), vt.end(), cmp1);
    for(int i = 0; i < n; i++) {
        if(book[vt[i].age] < 100) {
            v.push_back(vt[i]);
            book[vt[i].age]++;
        }
    }
    for(int i = 0; i < k; i++) {
        scanf("%d %d %d", &num, &amin, &amax);
        vector<node> t;
        for(int j = 0; j < v.size(); j++) {
            if(v[j].age >= amin && v[j].age <= amax)
                t.push_back(v[j]);
        }
        if(i != 0) printf("\n");
        printf("Case #%d:", i + 1);
        int flag = 0;
        for(int j = 0; j < num && j < t.size(); j++) {
            printf("\n%s %d %d", t[j].name, t[j].age, t[j].money);
            flag = 1;
        }
        if(flag == 0) printf("\nNone");
    }
    return 0;
}

本文來自部落格園，作者：勇往直前的力量，轉載請註明原文連結：https://www.cnblogs.com/moonlight1999/p/15585186.html