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 Algebra), 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
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.
Output
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".
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
排名不应该只是简单的排序算法：
比如说 100，100，99，…… 得99分是人是 第二名，而不是第三名。
所以，应该先对分数进行排序，然后在更新名次，算法如下：
1 SS[0].ra=1； //第一个肯定是第一名
2
3 int count=1;
4
5 for(i=1;i<n;i++)
6
7 {
8
9 if(SS[i].A==SS[i-1].A)//A为平均分
10
11 {
12
13 SS[i].ra=SS[i-1].ra;//ra 平均分排名
14
15 count++;//如果分数相同，那么名次和前一个相同，计数器++
16
17 }
18
19 else
20
21 {
22
23 SS[i].ra=SS[i-1].ra+count;
24
25 count=1;//计算器记得还原
26
27 }
28
29 }
AC代码：
1 #include <iostream>
2
3 #include <algorithm>
4
5 #include <string>
6
7 using namespace std;
8
9
10
11 struct stu
12
13 {
14
15 string ID;
16
17 int A,C,M,E;
18
19 int ra,rc,rm,re;
20
21 };
22
23
24
25 stu SS[10001];
26
27
28
29
30
31
32
33 bool cmp1(stu a,stu b)
34
35 {
36
37 return a.A>b.A;
38
39 }
40
41
42
43
44
45 bool cmp2(stu a,stu b)
46
47 {
48
49 return a.C>b.C;
50
51 }
52
53
54
55 bool cmp3(stu a,stu b)
56
57 {
58
59 return a.M>b.M;
60
61 }
62
63
64
65
66
67 bool cmp4(stu a,stu b)
68
69 {
70
71 return a.E>b.E;
72
73 }
74
75
76
77 int main()
78
79 {
80
81
82
83 int n;
84
85 while(cin>>n)
86
87 {
88
89 int m,i;
90
91 cin>>m;
92
93
94
95 for(i=0;i<n;i++)
96
97 {
98
99
100
101 cin>>SS[i].ID>>SS[i].C>>SS[i].M>>SS[i].E;
102
103 }
104
105
106
107
108
109 for(i=0;i<n;i++)
110
111 {
112
113 SS[i].A=(SS[i].C+SS[i].M+SS[i].E)/3;
114
115 SS[i].ra=1;
116
117 SS[i].rc=1;
118
119 SS[i].rm=1;
120
121 SS[i].re=1;
122
123 }
124
125
126
127
128
129
130
131 sort(SS,SS+n,cmp1);
132
133 int count=1;
134
135 for(i=1;i<n;i++)
136
137 {
138
139 if(SS[i].A==SS[i-1].A)
140
141 {
142
143 SS[i].ra=SS[i-1].ra;
144
145 count++;
146
147 }
148
149 else
150
151 {
152
153 SS[i].ra=SS[i-1].ra+count;
154
155 count=1;
156
157 }
158
159 }
160
161
162
163
164
165 sort(SS,SS+n,cmp2);
166
167
168
169 count=1;
170
171 for(i=1;i<n;i++)
172
173 {
174
175 if(SS[i].C==SS[i-1].C)
176
177 {
178
179 SS[i].rc=SS[i-1].rc;
180
181 count++;
182
183 }
184
185 else
186
187 {
188
189 SS[i].rc=SS[i-1].rc+count;
190
191 count=1;
192
193 }
194
195 }
196
197
198
199
200
201 sort(SS,SS+n,cmp3);
202
203 count=1;
204
205 for(i=1;i<n;i++)
206
207 {
208
209 if(SS[i].M==SS[i-1].M)
210
211 {
212
213 SS[i].rm=SS[i-1].rm;
214
215 count++;
216
217 }
218
219 else
220
221 {
222
223 SS[i].rm=SS[i-1].rm+count;
224
225 count=1;
226
227 }
228
229 }
230
231
232
233
234
235 sort(SS,SS+n,cmp4);
236
237 count=1;
238
239 for(i=1;i<n;i++)
240
241 {
242
243 if(SS[i].E==SS[i-1].E)
244
245 {
246
247 SS[i].re=SS[i-1].re;
248
249 count++;
250
251 }
252
253 else
254
255 {
256
257 SS[i].re=SS[i-1].re+count;
258
259 count=1;
260
261 }
262
263 }
264
265
266
267
268
269 for(i=0;i<m;i++)
270
271 {
272
273 string goal;int j;
274
275 cin>>goal;
276
277 for(j=0;j<n;j++)
278
279 if(SS[j].ID==goal) break;
280
281
282
283 if(j==n) cout<<"N/A"<<endl;
284
285 else{
286
287 char high=‘E‘;
288
289 int temp=SS[j].re;
290
291 if(temp>=SS[j].rm)
292
293 {
294
295 high=‘M‘;
296
297 temp=SS[j].rm;
298
299 }
300
301 if(temp>=SS[j].rc)
302
303 {
304
305 high=‘C‘;
306
307 temp=SS[j].rc;
308
309 }
310
311 if(temp>=SS[j].ra)
312
313 {
314
315 high=‘A‘;
316
317 temp=SS[j].ra;
318
319 }
320
321
322
323 cout<<temp<<" "<<high<<endl;
324
325 }
326
327 }
328
329 }
330
331 return 0;
332
333 }
334
335