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題目
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.
Input Specification:
Each input file contains one test case. Each case starts with a line
containing 2 numbers and
(
), which are the total number of
students, and the number of students who would check their ranks,
respectively. Then 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
lines, each containing a student ID.
Output Specification:
For each of the 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.
Sample Input:
5 6
310101 98 85 88
310102 70 95 88
310103 82 87 94
310104 91 91 91
310105 85 90 90
310101
310102
310103
310104
310105
999999
Sample Output:
1 C
1 M
1 E
1 A
3 A
N/A
思路
計算排名的方法：
很多人用的結(jié)構(gòu)體排序，這樣是比較耗時的（尤其是基數(shù)很大的時候）。
我的方法是在讀取學(xué)生成績的時候順便記錄0~100分各有多少人，然后根據(jù)得分人數(shù)計算排名。舉個例子：
- 如果100分有1人，99分有2人，98分有3人，97分有4人。
那么
- 得100分的排名是1，得99分的排名是2（并列），98分的排名是4（并列），97分的排名是7（并列）。
規(guī)律是什么？如果學(xué)生獲得的分數(shù)分別有a1>a2>a3>...>ai>...，那么
- 分數(shù)a[i]的名次 = 分數(shù)a[i-1]的名次 + 分數(shù)a[i-1]的人數(shù)
- 分數(shù)a[1]的名次 = 1
這其實是遞歸定義，如果定義一個不存在的高分a0（比如101分）人數(shù)=1（代碼中一開始的初始化就是做這個），那么上面的關(guān)系就簡化成了
- 分數(shù)a[i]的名次 = a[0]~a[i-1]的人數(shù)總和，(i>0)
可以驗證一下：97分的排名=98~101分人數(shù)總和=3+2+1+1=7
平均分計算方法：
仔細看題目中給的例子，平均分是四舍五入的，不過我試過在算平均分的時候直接3個數(shù)相加除以3，還是能AC。所以要不是題目沒有設(shè)置這樣的測試點，要不就是題目認為兩種算法結(jié)果都正確。
代碼
最新代碼@github，歡迎交流
#include <stdio.h>
typedef struct Student{ int ID, score[4]; } Student;
int main()
{
/* scores array stores the number of students of any score */
int N, M, ID, scores[4][102] = {{0}};
/* Setting a score of 101 is to make it easy to calculate rank */
scores[0][101] = scores[1][101] = scores[2][101] = scores[3][101] = 1;
Student students[2000];
scanf("%d %d", &N, &M);
for(int i = 0; i < N; i++)
{ /* record all the scores for every student: A, C, M, E */
Student *s = students + i;
scanf("%d %d %d %d", &s->ID, &s->score[1], &s->score[2], &s->score[3]);
s->score[0] = (s->score[1] + s->score[2] + s->score[3] + 1) / 3;
for(int j = 0; j < 4; j++) /* +1 for rounding */
scores[j][s->score[j]]++; /* record how many got this score */
}
for(int i = 0; i < M; i++)
{
int max = 3, ranks[4] = {0}, stu;
scanf("%d", &ID);
for(stu = 0; stu < N && students[stu].ID != ID; stu++) ; /* find */
if(stu == N) { puts("N/A"); continue;}
for(int j = 3; j >= 0; j--)
{ /* calculate the rank: sum the counts from score + 1 to 101 */
/* e.g. score 100 will have rank 1, since scores[101] is 1 */
for(int score = 100; score >= students[stu].score[j]; score--)
ranks[j] += scores[j][score + 1];
if(ranks[j] <= ranks[max]) /* The best rank with highest priority */
max = j;
}
printf("%d %c\n", ranks[max], "ACME"[max]);
}
}
