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:
N
name[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:
4
Tom CS000001 59
Joe Math990112 89
Mike CS991301 100
Mary EE990830 95
60 100
Sample Output 1:
Mike CS991301
Mary EE990830
Joe Math990112
Sample Input 2:
2
Jean AA980920 60
Ann CS01 80
90 95
Sample Output 2:
NONE
简单排序
#include<cstdio>
#include<stack>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn = 1e5 + 10;
char name[maxn][25];
char id[maxn][25];
int score[maxn];
int n, l, r;
vector<int> ans;
bool cmp(const int &x, const int&y)
{
return score[x] > score[y];
}
int main()
{
scanf("%d", &n);
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;
}