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PAT Ranking (排名)

系統 1605 0

PAT Ranking (排名)

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Programming Ability Test (PAT) is organized by the College of Computer Science and Technology of Zhejiang University. Each test is supposed to run simultaneously in several places, and the ranklists will be merged immediately after the test. Now it is your job to write a program to correctly merge all the ranklists and generate the final rank.

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?Input Specification:

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Each input file contains one test case. For each case, the first line contains a positive number N (<=100), the number of test locations. Then N ranklists follow, each starts with a line containing a positive integer K (<=300), the number of testees, and then K lines containing the registration number (a 13-digit number) and the total score of each testee. All the numbers in a line are separated by a space.

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?Output Specification:

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For each test case, first print in one line the total number of testees. Then print the final ranklist in the following format:

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registration_number final_rank location_number local_rank

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The locations are numbered from 1 to N. The output must be sorted in nondecreasing order of the final ranks. The testees with the same score must have the same rank, and the output must be sorted in nondecreasing order of their registration numbers.

Sample Input:

2

5

1234567890001 95

1234567890005 100

1234567890003 95

1234567890002 77

1234567890004 85

4

1234567890013 65

1234567890011 25

1234567890014 100

1234567890012 85

?Sample Output:

9

1234567890005 1 1 1

1234567890014 1 2 1

1234567890001 3 1 2

1234567890003 3 1 2

1234567890004 5 1 4

1234567890012 5 2 2

1234567890002 7 1 5

1234567890013 8 2 3

1234567890011 9 2 4

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先分組排名,再全部排名

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            1
        
         #include <iostream>


        
            2
        
        
            3
        
         #include <
        
          string
        
        >


        
            4
        
        
            5
        
         #include <algorithm>


        
            6
        
        
            7
        
         #include <vector>


        
            8
        
        
            9
        
        
          using
        
        
          namespace
        
        
           std;


        
        
           10
        
        
           11
        
        
           12
        
        
           13
        
        
          struct
        
        
           stu


        
        
           14
        
        
           15
        
        
          {


        
        
           16
        
        
           17
        
        
          int
        
        
           loc,frank,lrank,gra;


        
        
           18
        
        
           19
        
        
          string
        
        
           name;


        
        
           20
        
        
           21
        
        
          };


        
        
           22
        
        
           23
        
        
           24
        
        
           25
        
        
           26
        
        
           27
        
        
          bool
        
        
           cmp(stu s1,stu s2)


        
        
           28
        
        
           29
        
        
          {


        
        
           30
        
        
           31
        
        
          if
        
        (s1.frank==
        
          s2.frank)


        
        
           32
        
        
           33
        
        
          return
        
         s1.name<
        
          s2.name;


        
        
           34
        
        
           35
        
        
           36
        
        
           37
        
        
          return
        
         s1.frank<
        
          s2.frank;


        
        
           38
        
        
           39
        
        
          }


        
        
           40
        
        
           41
        
        
           42
        
        
           43
        
        
          bool
        
        
           cmpg(stu s1,stu s2)


        
        
           44
        
        
           45
        
        
          {


        
        
           46
        
        
           47
        
        
           48
        
        
           49
        
        
          return
        
         s1.gra>
        
          s2.gra;


        
        
           50
        
        
           51
        
        
          }


        
        
           52
        
        
           53
        
        
           54
        
        
           55
        
        
          int
        
        
           main()


        
        
           56
        
        
           57
        
        
          {


        
        
           58
        
        
           59
        
        
           60
        
        
           61
        
        
          int
        
         n;
        
          int
        
         k;
        
          int
        
        
           rank,count;


        
        
           62
        
        
           63
        
        
          int
        
        
           i,j;


        
        
           64
        
        
           65
        
        
          while
        
        (cin>>
        
          n)


        
        
           66
        
        
           67
        
        
            {


        
        
           68
        
        
           69
        
               vector<stu>
        
           total;


        
        
           70
        
        
           71
        
        
          for
        
        (i=
        
          1
        
        ;i<=n;i++
        
          )


        
        
           72
        
        
           73
        
        
               {


        
        
           74
        
        
           75
        
                  cin>>
        
          k;


        
        
           76
        
        
           77
        
                     vector<stu>
        
           s(k);


        
        
           78
        
        
           79
        
        
          for
        
        (j=
        
          0
        
        ;j<k;j++
        
          )


        
        
           80
        
        
           81
        
        
                      {


        
        
           82
        
        
           83
        
                        cin>>s[j].name>>
        
          s[j].gra;


        
        
           84
        
        
           85
        
                        s[j].loc=
        
          i;


        
        
           86
        
        
           87
        
        
                      }


        
        
           88
        
        
           89
        
        
                      sort(s.begin(),s.end(),cmpg);


        
        
           90
        
        
           91
        
                     count=
        
          1
        
        
          ;


        
        
           92
        
        
           93
        
                     s[
        
          0
        
        ].lrank=
        
          1
        
        
          ;


        
        
           94
        
        
           95
        
                     total.push_back(s[
        
          0
        
        
          ]);


        
        
           96
        
        
           97
        
        
          for
        
        (j=
        
          0
        
        ;j<k-
        
          1
        
        ;j++
        
          )


        
        
           98
        
        
           99
        
        
                      {


        
        
          100
        
        
          101
        
        
          if
        
        (s[j].gra==s[j+
        
          1
        
        
          ].gra)


        
        
          102
        
        
          103
        
        
                            {


        
        
          104
        
        
          105
        
                                 s[j+
        
          1
        
        ].lrank=
        
          s[j].lrank;


        
        
          106
        
        
          107
        
                                 count++
        
          ;


        
        
          108
        
        
          109
        
        
                            }


        
        
          110
        
        
          111
        
        
          else
        
        
          112
        
        
          113
        
        
                            {


        
        
          114
        
        
          115
        
                               s[j+
        
          1
        
        ].lrank=s[j].lrank+
        
          count;


        
        
          116
        
        
          117
        
                                 count=
        
          1
        
        
          ;


        
        
          118
        
        
          119
        
        
                            }


        
        
          120
        
        
          121
        
                           total.push_back(s[j+
        
          1
        
        
          ]);


        
        
          122
        
        
          123
        
        
                      }


        
        
          124
        
        
          125
        
        
               }


        
        
          126
        
        
          127
        
        
          128
        
        
          129
        
        
                    sort(total.begin(),total.end(),cmpg);


        
        
          130
        
        
          131
        
                   count=
        
          1
        
        
          ;


        
        
          132
        
        
          133
        
                     total[
        
          0
        
        ].frank=
        
          1
        
        
          ;


        
        
          134
        
        
          135
        
        
          for
        
        (j=
        
          0
        
        ;j<total.size()-
        
          1
        
        ;j++
        
          )


        
        
          136
        
        
          137
        
        
                      {


        
        
          138
        
        
          139
        
        
          if
        
        (total[j].gra==total[j+
        
          1
        
        
          ].gra)


        
        
          140
        
        
          141
        
        
                            {


        
        
          142
        
        
          143
        
                                 total[j+
        
          1
        
        ].frank=
        
          total[j].frank;


        
        
          144
        
        
          145
        
                                 count++
        
          ;


        
        
          146
        
        
          147
        
        
                            }


        
        
          148
        
        
          149
        
        
          else
        
        
          150
        
        
          151
        
        
                            {


        
        
          152
        
        
          153
        
                               total[j+
        
          1
        
        ].frank=total[j].frank+
        
          count;


        
        
          154
        
        
          155
        
                                 count=
        
          1
        
        
          ;


        
        
          156
        
        
          157
        
        
                            }


        
        
          158
        
        
          159
        
        
                      }


        
        
          160
        
        
          161
        
        
          162
        
        
          163
        
        
                      sort(total.begin(),total.end(),cmp);


        
        
          164
        
        
          165
        
                     cout<<total.size()<<
        
          endl;


        
        
          166
        
        
          167
        
        
          for
        
        (i=
        
          0
        
        ;i<total.size();i++
        
          )


        
        
          168
        
        
          169
        
                           cout<<total[i].name<<
        
          "
        
        
          "
        
        <<total[i].frank<<
        
          "
        
        
          "
        
        <<total[i].loc<<
        
          "
        
        
          "
        
        <<total[i].lrank<<
        
          endl;


        
        
          170
        
        
          171
        
        
          172
        
        
          173
        
        
            }


        
        
          174
        
        
          175
        
        
          return
        
        
          0
        
        
          ;


        
        
          176
        
        
          177
        
        
          }


        
        
          178
        
        
          179
        
      
View Code

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PAT Ranking (排名)


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