PAT 甲级 1055. The World's Richest (25)
来源:互联网 发布:卡司数据 编辑:程序博客网 时间:2024/06/05 04:24
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_WorthThe 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 4Zoe_Bill 35 2333Bob_Volk 24 5888Anny_Cin 95 999999Williams 30 -22Cindy 76 76000Alice 18 88888Joe_Mike 32 3222Michael 5 300000Rosemary 40 5888Dobby 24 5888Billy 24 5888Nobody 5 04 15 454 30 354 5 951 45 50Sample Output:
Case #1:Alice 18 88888Billy 24 5888Bob_Volk 24 5888Dobby 24 5888Case #2:Joe_Mike 32 3222Zoe_Bill 35 2333Williams 30 -22Case #3:Anny_Cin 95 999999Michael 5 300000Alice 18 88888Cindy 76 76000Case #4:None
#include <iostream>#include <vector>#include <algorithm>#include <cstring>#include <stack>using namespace std;struct node {char name[10];int age;int 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;elsereturn(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;}
- 1055. The World's Richest (25)-PAT甲级真题
- 【PAT甲级】1055. The World's Richest (25)
- PAT(甲级)1055. The World's Richest (25)
- 1055. The World's Richest (25) PAT甲级
- PAT甲级练习1055. The World's Richest (25)
- PAT甲级1055. The World's Richest (25)
- PAT 甲级 1055. The World's Richest (25)
- PAT甲级 1055. The World's Richest (25)
- 【PAT】1055. The World's Richest (25)
- pat 1055. The World's Richest (25)
- pat 1055. The World's Richest (25)
- PAT: 1055. The World's Richest (25)
- PAT 1055. The World's Richest (25)
- pat 1055. The World's Richest (25)
- PAT 1055. The World's Richest (25)
- PAT 1055. The World's Richest (25)
- <PAT>1055. The World's Richest
- PAT 1055. The World's Richest
- eclipse中Tomcat启动时报错Cannot assign requested address
- Java图片处理开源框架
- 数组的原地排序和保护排序
- 适配器设计模式
- 栈和队列---最大值减去最小值小于或等于num的子数组数量
- PAT 甲级 1055. The World's Richest (25)
- 敏捷转型中常见的问题
- 趣味回顾会-回顾
- 趣味回顾会-鼓舞团队热情
- 如何进行高效迅速的CodeReview
- 趣味回顾会之Check in
- AMM敏捷成熟度评估框架介绍
- 中兴通讯某分组产品敏捷转型实践
- 丰田生产系统TPS与敏捷实践