PAT 1025 PAT Ranking

1025 PAT Ranking

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.

Input Specification:

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.

Output Specification:

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

registration_number final_rank location_number local_rank

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

总结：这道题不难，根据前面做过的几个差不多的结构体排序的问题，可以用差不多的方法做来，让我没有想到的是这次调试很快就调对了（很少T...T），还是挺高兴的，所以多练还是有用的！

基本思路：在一开始输入的时候将每个地区的人排名一遍，然后根据分数赋上排名，最后总的排名一边，如果分数相等，那就按序号的高低排名（从低到高） 

#include 
#include 
using namespace std;
 
const int N=30010;
struct Stu{
    string s;
    int score;
    int rank[3];//rank[0]表示全部排名，rank[1]表示地区，rank[2]表示当地排名
}stu[N];
 
bool cmp(Stu a,Stu b){
    return a.score>b.score;
}
 
bool cmp2(Stu a,Stu b){
    if(a.score!=b.score)    return a.score>b.score;
    else return a.s
}
 
int main(){
    int n,k,sum=0,index=0;
    cin >> n;
    
    for(int i=0;i
        scanf("%d",&k);
        for(int j=0;j
            cin >> stu[index].s >> stu[index].score;
            stu[index++].rank[1]=i+1;
        }
        sort(stu+sum,stu+sum+k,cmp);
        stu[sum].rank[2]=1;
        for(int j=1;j
            stu[sum+j].rank[2]=j+1;
            if(stu[sum+j].score==stu[sum+j-1].score)
                stu[sum+j].rank[2]=stu[sum+j-1].rank[2];
        }
        sum+=k;
    }
    
    sort(stu,stu+sum,cmp2);
    stu[0].rank[0]=1;
    for(int j=1;j
        stu[j].rank[0]=j+1;
        if(stu[j].score==stu[j-1].score)
            stu[j].rank[0]=stu[j-1].rank[0];
    }
    
    printf("%d\n",sum);
    for(int i=0;i
        cout << stu[i].s << ' ' << stu[i].rank[0] << ' ' << stu[i].rank[1] << ' ' << stu[i].rank[2] << endl;
    }
    
    return 0;
}

依照惯例，还是看看大佬的！

还是简洁明了，通俗易懂！(没有用cin cout 耗时少了很多)

#include 
#include 
#include 
using namespace std;
struct student {
    long long int no;
    int score, finrank, loca, locarank;
};
bool cmp1(student a, student b) {
    return a.score != b.score ? a.score > b.score : a.no < b.no;
}
int main() {
    int n, m;
    scanf("%d", &n);
    vector fin;
    for(int i = 1; i <= n; i++) {
        scanf("%d", &m);
        vector v(m);
        for(int j = 0; j < m; j++) {
            scanf("%lld %d", &v[j].no, &v[j].score);
            v[j].loca = i;
        }
        sort(v.begin(), v.end(), cmp1);
        v[0].locarank = 1;
        fin.push_back(v[0]);
        for(int j = 1; j < m; j++) {
            v[j].locarank = (v[j].score == v[j - 1].score) ? (v[j - 1].locarank) : (j + 1);
            fin.push_back(v[j]);
        }
    }
    sort(fin.begin(), fin.end(), cmp1);
    fin[0].finrank = 1;
    for(int j = 1; j < fin.size(); j++)
        fin[j].finrank = (fin[j].score == fin[j - 1].score) ? (fin[j - 1].finrank) : (j + 1);
    printf("%d\n", fin.size());
    for(int i = 0; i < fin.size(); i++)
        printf("%013lld %d %d %d\n", fin[i].no, fin[i].finrank, fin[i].loca, fin[i].locarank);
    return 0;
}

好好学习，天天向上！

我要考研！

2022.11.2

//思路很简单，在输入每一个地区的成绩时将地区成绩排名，然后把地区合并和在一个数组中
//最后将数组中所有参赛者的成绩进行排名，打印结果即可
#include 
#include 
#include 
using namespace std;
 
struct node{
    string id;
    int f,loc,locrank,rank;
};
//获取排名
bool cmp1(node a,node b){
    if(a.f!=b.f)    return a.f>b.f;
    return a.id
}
int main(){
    int n,k,cot=0;
    scanf("%d",&n);
    vector v[n];
    
    for(int i=0;i
        scanf("%d",&k);
        cot+=k;
        string id;
        int f;
        v[i].resize(k);
        for(int j=0;j
            cin >> id >> f;
            v[i][j]={id,f,i+1};
        }
        sort(v[i].begin(),v[i].end(),cmp1);
        int pre=-1,rank=1;
        for(int j=0;j
            if(pre!=v[i][j].f)  rank=j+1;
            v[i][j].locrank=rank;
            pre=v[i][j].f;
        }
    }
    vector s(300010);
    int t=0,pre=-1,rank=1;
    for(int i=0;i
        for(int j=0;jsize();j++)
            s[t++]=v[i][j];
    }
    printf("%d\n",t);
    sort(s.begin(),s.begin()+t,cmp1);
    for(int i=0;isize();i++){
        if(pre!=s[i].f) rank=i+1;
        s[i].rank=rank;
        pre=s[i].f;
    }
    for(int i=0;i
        cout << s[i].id;
        printf(" %d %d %d\n",s[i].rank,s[i].loc,s[i].locrank);
    }
    return 0;
}