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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.
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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
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?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.
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Input
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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.
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Output
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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.
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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.
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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
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排名不應該只是簡單的排序算法：
比如說??100，100，99，……???得99分是人是 第二名，而不是第三名。
所以，應該先對分數進行排序，然后在更新名次，算法如下：
????????
? ??
1
SS[
0
].ra=
1
；
//
第一個肯定是第一名
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為平均分
10
11
{
12
13
SS[i].ra=SS[i-
1
].ra;
//
ra 平均分排名
14
15
count++;
//
如果分數相同，那么名次和前一個相同，計數器++
16
17
}
18
19
else
20
21
{
22
23
SS[i].ra=SS[i-
1
].ra+
count;
24
25
count=
1
;
//
計算器記得還原
26
27
}
28
29
}
?
?
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AC代碼：
?
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
?
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