PAT (Advanced Level) Practise 1083 List Grades (25)

1083. List Grades (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Given a list of N student records with name, ID and grade. You are supposed to sort the records with respect to the grade in non-increasing order, and output those student records of which the grades are in a given interval.

Input Specification:

Each input file contains one test case. Each case is given in the following format:

Nname[1] ID[1] grade[1]name[2] ID[2] grade[2]... ...name[N] ID[N] grade[N]grade1 grade2

where name[i] and ID[i] are strings of no more than 10 characters with no space, grade[i] is an integer in [0, 100], grade1 and grade2 are the boundaries of the grade's interval. It is guaranteed that all the grades are distinct.

Output Specification:

For each test case you should output the student records of which the grades are in the given interval [grade1, grade2] and are in non-increasing order. Each student record occupies a line with the student's name and ID, separated by one space. If there is no student's grade in that interval, output "NONE" instead.

Sample Input 1:

4Tom CS000001 59Joe Math990112 89Mike CS991301 100Mary EE990830 9560 100

Sample Output 1:

Mike CS991301Mary EE990830Joe Math990112

Sample Input 2:

2Jean AA980920 60Ann CS01 8090 95

Sample Output 2:

NONE

题意：给出n个人的名字，id和分数，问在一个分数段里的有哪些人，按字典序输出

#include <iostream>    #include <cstdio>    #include <string>    #include <cstring>    #include <algorithm>    #include <cmath>    #include <vector>    #include <map>    #include <set>    #include <queue>    #include <stack>    #include <functional>    #include <climits>    using namespace std;#define LL long long    const int INF = 0x7FFFFFFF;char name[100009][25],id[100009][25];int score[100009];int n, l, r;vector<int> ans;bool cmp(const int &x, const int&y){return score[x] > score[y];}int main(){while (~scanf("%d", &n)){ans.clear();for (int i = 0; i < n; i++) scanf("%s%s%d", name[i], id[i], &score[i]);scanf("%d%d", &l, &r);for (int i = 0; i < n; i++)if (score[i] >= l&&score[i] <= r) ans.push_back(i);sort(ans.begin(), ans.end(), cmp);if (ans.size()){for (int i = 0; i < ans.size(); i++)printf("%s %s\n", name[ans[i]], id[ans[i]]);}else printf("NONE\n");}return 0;}