PAT 甲级 1012 The Best Rank (25 分)
题目描述
To evaluate the performance of our first year CS majored students, we consider their grades of three courses only: C - C Programming Language, M - Mathematics (Calculus or Linear Algrbra), and E - English. At the mean time, we encourage students by emphasizing on their best ranks – that is, among the four ranks with respect to the three courses and the average grade, we print the best rank for each student.
For example, The grades of C, M, E and A - Average of 4 students are given as the following:
StudentID C M E A
310101 98 85 88 90
310102 70 95 88 84
310103 82 87 94 88
310104 91 91 91 91
Then the best ranks for all the students are No.1 since the 1st one has done the best in C Programming Language, while the 2nd one in Mathematics, the 3rd one in English, and the last one in average.
输入
Each input file contains one test case. Each case starts with a line containing 2 numbers N and M (≤2000), which are the total number of students, and the number of students who would check their ranks, respectively. Then N lines follow, each contains a student ID which is a string of 6 digits, followed by the three integer grades (in the range of [0, 100]) of that student in the order of C, M and E. Then there are M lines, each containing a student ID.
输出
For each of the M students, print in one line the best rank for him/her, and the symbol of the corresponding rank, separated by a space.
The priorities of the ranking methods are ordered as A > C > M > E. Hence if there are two or more ways for a student to obtain the same best rank, output the one with the highest priority.
If a student is not on the grading list, simply output N/A.
思路
这道题其实很简单，我们对每一个部分排一次序列，然后找到对应的位置即可。
代码
#include<iostream>
#include<cstdio>
#include<stdlib.h>
#include<string.h>
#include<algorithm>
using namespace std;
int N, M;
typedef struct node {
char id[10] = {
0 };
int C;
int M;
int E;
int A;
}Grade;
Grade cj[20000];
typedef struct Node {
char id[10] = {
0 };
int rank;
char ch;
}Ans;
Ans ans[20000];
bool cmp1(Grade a, Grade b)
{
return a.A > b.A;
}
bool cmp2(Grade a, Grade b)
{
return a.C > b.C;
}
bool cmp3(Grade a, Grade b)
{
return a.M > b.M;
}
bool cmp4(Grade a, Grade b)
{
return a.E > b.E;
}
int check(int n, char c)
{
int i;
if (c == 'A')
for (i = n - 1; i >= 0; i--)
{
if (cj[n].A != cj[i].A)
{
break;
}
}
if (c == 'C')
for (i = n - 1; i >= 0; i--)
{
if (cj[n].C != cj[i].C)
{
break;
}
}
if (c == 'M')
for (i = n - 1; i >= 0; i--)
{
if (cj[n].M != cj[i].M)
{
break;
}
}
if (c == 'E')
for (i = n - 1; i >= 0; i--)
{
if (cj[n].E != cj[i].E)
{
break;
}
}
return i + 1;
}
void change(char c)
{
for (int i = 0; i < N; i++)
{
for (int j = 0; j < N; j++)
{
if (strcmp(cj[i].id, ans[j].id) == 0)
{
int temp_rank = check(i, c);
if (temp_rank + 1 < ans[j].rank)
{
ans[j].ch = c;
ans[j].rank = temp_rank + 1;
}
break;
}
}
}
}
int main()
{
scanf("%d%d", &N, &M);
for (int i = 0; i < N; i++)
{
scanf("%s %d %d %d", cj[i].id, &cj[i].C, &cj[i].M, &cj[i].E);
cj[i].A = (cj[i].C + cj[i].M + cj[i].E) / 3;
}
sort(cj, cj + N, cmp1);
for (int i = 0; i < N; i++)
{
strcpy(ans[i].id, cj[i].id);
ans[i].rank = check(i, 'A') + 1;
ans[i].ch = 'A';
}
sort(cj, cj + N, cmp2);
change('C');
sort(cj, cj + N, cmp3);
change('M');
sort(cj, cj + N, cmp4);
change('E');
for (int i = 0; i < M; i++)
{
char temp[10];
scanf("%s", temp);
int flag = 0;
for (int j = 0; j < N; j++)
{
if (strcmp(temp, ans[j].id) == 0)
{
flag = 1;
printf("%d %c\n", ans[j].rank, ans[j].ch);
break;
}
}
if (flag == 0)
{
printf("N/A\n");
}
}
}