pat a1025

最新推荐文章于 2024-09-03 22:15:15 发布

原创最新推荐文章于 2024-09-03 22:15:15 发布 · 181 阅读

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本文介绍了一个用于合并多个考试地点的编程能力测试(PAT)排名列表的算法。该算法需要处理来自不同地点的输入数据，每个地点包含考生的注册号和总分，然后生成一个全局排名列表，考虑到相同分数的考生排名相同的情况。

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

代码
#include<stdio.h>
#include
using namespace std;
struct stu
{
int fen;
long long id;
int mingci[2]={1,1};
int location;
}st[30000];
bool cmp(stu a,stu b)
{
if(a.fen!=b.fen)
return a.fen>b.fen;
else if(a.id!=b.id)
return a.id<b.id;
}
int main()
{
bool cmp(stu a,stu b);
int n;
int t=0;
int k[101]={0};
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%d",&k[i]);
for(int j=0;j<k[i];j++)
{
scanf("%lld%d",&st[t].id,&st[t].fen);
st[t].location=i;
t++;
}
sort(st+t-k[i],st+t,cmp);
for(int j=t-k[i]+1;j<t;j++)
{
if(st[j].fenst[j-1].fen)
st[j].mingci[0]=st[j-1].mingci[0];
else
st[j].mingci[0]=j-t+k[i]+1;
}
}
sort(st,st+t,cmp);
for(int j=1;j<t;j++)
{
if(st[j].fenst[j-1].fen)
st[j].mingci[1]=st[j-1].mingci[1];
else
st[j].mingci[1]=j+1;
}