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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.
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
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原文地址：http://www.cnblogs.com/xiaoyesoso/p/4261443.html
