10-排序5 PAT Judge
用了冒泡和插入排序 果然没有什么本质区别。。都是运行超时
用库函数sort也超时
The ranklist of PAT is generated from the status list, which shows the scores of the submittions. This time you are supposed to generate the ranklist for PAT.
Input Specification:
Each input file contains one test case. For each case, the first line contains 3 positive integers, N (≤10^4), the total number of users, K (≤5), the total number of problems, and M ≤10^5), the total number of submittions. It is then assumed that the user id's are 5-digit numbers from 00001 to N, and the problem id's are from 1 to K. The next line contains K positive integers p[i] (i=1, ..., K), where p[i] corresponds to the full mark of the i-th problem. Then M lines follow, each gives the information of a submittion in the following format:
user_id problem_id partial_score_obtained
where partial_score_obtained is either -1 if the submittion cannot even pass the compiler, or is an integer in the range [0, p[problem_id]]. All the numbers in a line are separated by a space.
Output Specification:
For each test case, you are supposed to output the ranklist in the following format:
rank user_id total_score s[1] ... s[K]
where rank is calculated according to the total_score, and all the users with the same total_score obtain the same rank; and s[i] is the partial score obtained for the i-th problem. If a user has never submitted a solution for a problem, then "-" must be printed at the corresponding position. If a user has submitted several solutions to solve one problem, then the highest score will be counted.
The ranklist must be printed in non-decreasing order of the ranks. For those who have the same rank, users must be sorted in nonincreasing order according to the number of perfectly solved problems. And if there is still a tie, then they must be printed in increasing order of their id's. For those who has never submitted any solution that can pass the compiler, or has never submitted any solution, they must NOT be shown on the ranklist. It is guaranteed that at least one user can be shown on the ranklist.
Sample Input:
7 4 20
20 25 25 30
00002 2 12
00007 4 17
00005 1 19
00007 2 25
00005 1 20
00002 2 2
00005 1 15
00001 1 18
00004 3 25
00002 2 25
00005 3 22
00006 4 -1
00001 2 18
00002 1 20
00004 1 15
00002 4 18
00001 3 4
00001 4 2
00005 2 -1
00004 2 0
Sample Output:
1 00002 63 20 25 - 18
2 00005 42 20 0 22 -
2 00007 42 - 25 - 17
2 00001 42 18 18 4 2
5 00004 40 15 0 25 -
#include <stdio.h> struct user
{
int id;
int score[];
int solved;
int sum;
int flag;
}users[]; int main()
{
int N, K, M, max = ;
scanf("%d %d %d", &N, &K, &M);
int p[] = {};
for(int i = ; i <= K; i++) {
scanf("%d",&p[i]);
max += p[i];
} //初始化
for(int i = ; i <= N;i++) {
users[i].id = i;
users[i].solved = ;
users[i].sum = ;
users[i].flag = ; for(int j = ; j < ; j++)
users[i].score[j] = -;
} int tmpUserId, tmpProId, tmpPSO;//user_id problem_id partial_score_obtained
for(int i = ; i < M; i++) {
scanf("%d %d %d", &tmpUserId, &tmpProId, &tmpPSO);
if(users[tmpUserId].score[tmpProId] < tmpPSO) {
if(tmpPSO > -)
users[tmpUserId].flag = ; //有提交 how about -1
users[tmpUserId].score[tmpProId] = tmpPSO; //更新最高分 if(tmpPSO == p[tmpProId]) //满分的问题数
users[tmpUserId].solved++;
}
} //统计sum
for(int i = ; i <= N; i++) {
for(int j = ; j < ; j++) {
if(users[i].score[j] > )
users[i].sum += users[i].score[j];
}
}
// for(int i = 1; i <= N; i++)
// printf("%d : %d\n",i,users[i].sum);
//冒泡排序
int sort[]; //存放顺序
for(int i = ; i <= N; i++)
sort[i] = i;
for(int i = N; i > ; i--) {
int flag = ;
for(int j = ; j < i; j++) {
if( users[sort[j]].sum < users[sort[j+]].sum ) {
int tmp = sort[j]; //交换次序
sort[j] = sort[j+];
sort[j+] = tmp;
flag = ;
// printf("j:%d sort[j]:%d %d\n",j,sort[j],sort[j+1]);
} else if(users[sort[j]].sum == users[sort[j+]].sum) { //如果总分一样
if(users[sort[j]].solved < users[sort[j+]].solved) { //根据完美解决问题数排序
int tmp = sort[j]; //交换次序
sort[j] = sort[j+];
sort[j+] = tmp;
flag = ;
} //由于是冒泡排序稳定,id小的在前面 不需要再判断
}
}
// for(int i = 1; i <= N; i++)
// printf("%d ",sort[i]);
// printf("\n");
if(flag == ) break;
} // for(int i = 1; i <= N; i++) {
// printf("%05d %d ", users[i].id, users[i].sum);
// for(int j = 1; j < K; j++) {
// if(users[i].score[j] == -1)
// users[i].score[j] = 0;
// if(users[i].score[j] >= 0)
// printf("%d ",users[i].score[j]);
// else printf("- ");
// }
// if(users[i].score[K] >= 0)
// printf("%d\n",users[i].score[K]);
// else printf("-\n");
// } //显示输出
int rank = ;
int currentSum = max;
for(int i = ; i <= N; i++) {
if(users[sort[i]].flag == )
continue; if(users[sort[i]].sum == currentSum)
;
else
rank = i; currentSum = users[sort[i]].sum;
printf("%d %05d %d ", rank, users[sort[i]].id, currentSum); for(int j = ; j < K; j++) {
if(users[sort[i]].score[j] == -)
users[sort[i]].score[j] = ;
if(users[sort[i]].score[j] >= )
printf("%d ",users[sort[i]].score[j]);
else printf("- ");
}
if(users[sort[i]].score[K] >= )
printf("%d\n",users[sort[i]].score[K]);
else printf("-\n");
} return ;
}
Bubble Sort
#include <stdio.h> struct user
{
int id;
int score[];
int solved;
int sum;
int flag;
}users[]; int main()
{
int N, K, M, max = ;
scanf("%d %d %d", &N, &K, &M);
int p[] = {};
for(int i = ; i <= K; i++) {
scanf("%d",&p[i]);
max += p[i];
} //初始化
for(int i = ; i <= N;i++) {
users[i].id = i;
users[i].solved = ;
users[i].sum = ;
users[i].flag = ; for(int j = ; j < ; j++)
users[i].score[j] = -;
} int tmpUserId, tmpProId, tmpPSO;//user_id problem_id partial_score_obtained
for(int i = ; i < M; i++) {
scanf("%d %d %d", &tmpUserId, &tmpProId, &tmpPSO);
if(users[tmpUserId].score[tmpProId] < tmpPSO ) {
if(tmpPSO > -)
users[tmpUserId].flag = ; //有提交
users[tmpUserId].score[tmpProId] = tmpPSO; //更新最高分 // if(tmpPSO == p[tmpProId]) //满分的问题数
// users[tmpUserId].solved++;
}
} //统计sum
for(int i = ; i <= N; i++) {
for(int j = ; j < ; j++) {
if(users[i].score[j] > ) {
// users[i].flag = 1; //有提交
users[i].sum += users[i].score[j];
if(users[i].score[j] == p[j]) //满分的问题数
users[i].solved++;
}
}
} //插入排序
int sort[]; //存放顺序
for(int i = ; i <= N; i++)
sort[i] = i;
int j;
for(int i = ; i <= N; i++) {
int tmp = sort[i];
for(j = i; j > ; j--) {
if(users[sort[j]].sum > users[sort[j-]].sum) {
sort[j] = sort[j-];
sort[j-] = tmp;
} else if(users[sort[j]].sum == users[sort[j-]].sum) {
if(users[sort[j]].solved > users[sort[j-]].solved) {
sort[j] = sort[j-];
sort[j-] = tmp;
}
} else //由于是插入排序稳定,id小的在前面 不需要再判断
break;
}
// for(int k = 1; k <= N; k++)
// printf("%d ",sort[k]);
// printf("\n");
} //显示输出
int rank = ;
int currentSum = max;
for(int i = ; i <= N; i++) {
if(users[sort[i]].flag == )
continue; if(users[sort[i]].sum == currentSum)
;
else
rank = i; currentSum = users[sort[i]].sum;
printf("%d %05d %d ", rank, users[sort[i]].id, currentSum); for(int j = ; j < K; j++) {
if(users[sort[i]].score[j] == -)
users[sort[i]].score[j] = ;
if(users[sort[i]].score[j] >= )
printf("%d ",users[sort[i]].score[j]);
else printf("- ");
}
if(users[sort[i]].score[K] >= )
printf("%d\n",users[sort[i]].score[K]);
else printf("-\n");
} return ;
}
Insertion sort
#include <iostream>
#include <algorithm>
#include <vector>
#include <cstdio>
using namespace std; struct user
{
int id;
int score[];
int solved;
int sum;
int flag;
}users[]; /*cmp函数的返回值为true和false或1和0,
若为true/1,则sort()函数为升序排列,
若为false/0,则sort()函数为降序排列。
*/
bool cmp(user u1, user u2) {
if (u1.sum != u2.sum){
return u1.sum > u2.sum;
} else{
if (u1.solved != u2.solved) {
return u1.solved > u2.solved;
} else{
return u1.id < u2.id;
}
}
} int main()
{
int N, K, M, max = ;
scanf("%d %d %d", &N, &K, &M);
int p[] = {};
for(int i = ; i <= K; i++) {
scanf("%d",&p[i]);
max += p[i];
} //初始化
for(int i = ; i <= N;i++) {
users[i].id = i;
users[i].solved = ;
users[i].sum = ;
users[i].flag = ; for(int j = ; j < ; j++)
users[i].score[j] = -;
} int tmpUserId, tmpProId, tmpPSO;//user_id problem_id partial_score_obtained
for(int i = ; i < M; i++) {
scanf("%d %d %d", &tmpUserId, &tmpProId, &tmpPSO);
if(users[tmpUserId].score[tmpProId] < tmpPSO ) {
if(tmpPSO > -)
users[tmpUserId].flag = ; //有提交
users[tmpUserId].score[tmpProId] = tmpPSO; //更新最高分 // if(tmpPSO == p[tmpProId]) //满分的问题数
// users[tmpUserId].solved++;
}
} vector<user> vec;
//统计sum
for(int i = ; i <= N; i++) {
for(int j = ; j < ; j++) {
if(users[i].score[j] > ) {
// users[i].flag = 1; //有提交
users[i].sum += users[i].score[j];
if(users[i].score[j] == p[j]) //满分的问题数
users[i].solved++;
}
}
if(users[i].sum >= )
vec.push_back(users[i]);
} sort(vec.begin(),vec.end(),cmp); //显示输出
int rank = ;
int currentSum = max;
for(int i = ; i < N; i++) {
if(vec[i].flag == )
continue; if(vec[i].sum == currentSum)
;
else
rank = i+; currentSum = vec[i].sum;
printf("%d %05d %d ", rank, vec[i].id, currentSum); for(int j = ; j < K; j++) {
if(vec[i].score[j] == -)
vec[i].score[j] = ;
if(vec[i].score[j] >= )
printf("%d ",vec[i].score[j]);
else printf("- ");
}
if(vec[i].score[K] >= )
printf("%d\n",vec[i].score[K]);
else printf("-\n");
}
return ;
}
sort
10-排序5 PAT Judge的更多相关文章
- PAT 1075 PAT Judge[比较]
1075 PAT Judge (25 分) The ranklist of PAT is generated from the status list, which shows the scores ...
- PTA 10-排序5 PAT Judge (25分)
题目地址 https://pta.patest.cn/pta/test/16/exam/4/question/677 5-15 PAT Judge (25分) The ranklist of PA ...
- A1075 PAT Judge (25)(25 分)
A1075 PAT Judge (25)(25 分) The ranklist of PAT is generated from the status list, which shows the sc ...
- PAT 甲级 1075 PAT Judge (25分)(较简单,注意细节)
1075 PAT Judge (25分) The ranklist of PAT is generated from the status list, which shows the scores ...
- PAT_A1075#PAT Judge
Source: PAT A1075 PAT Judge (25 分) Description: The ranklist of PAT is generated from the status lis ...
- PAT A1075 PAT Judge (25 分)——结构体初始化,排序
The ranklist of PAT is generated from the status list, which shows the scores of the submissions. Th ...
- PAT甲题题解-1075. PAT Judge (25)-排序
相当于是模拟OJ评测,这里注意最后输出:1.那些所有提交结果都是-1的(即均未通过编译器的),或者从没有一次提交过的用户,不需要输出.2.提交结果为-1的题目,最后输出分数是03.某个题目从没有提交过 ...
- PAT Judge
原题连接:https://pta.patest.cn/pta/test/16/exam/4/question/677 题目如下: The ranklist of PAT is generated fr ...
- PATA1075 PAT Judge (25 分)
The ranklist of PAT is generated from the status list, which shows the scores of the submissions. Th ...
随机推荐
- gnu c语言中的?:的作用
#include <stdio.h> #include <stdlib.h> char * test() { return "abc" ?: "f ...
- Redis多机功能介绍
Redis多机功能目的:以单台Redis服务器过渡到多台Redis服务器 Redis单机在生产环境中存在的问题 1.内存容量不足 Redis使用内存来存书数据库中的数据,但是对于一台机器来说,硬件的内 ...
- VR就是下一个浪潮_2016 (GMGC) 全球移动游戏大会观后感
"VR就是下一个浪潮" --2016 (GMGC) 全球移动游戏大会观后感 早在2014年参会Unity举办的一年一度的金立方盛典大会,就初次体验了VR头盔设备,于是印象深刻 ...
- 【Unity Shaders】学习笔记——SurfaceShader(一)认识结构
[Unity Shaders]学习笔记——SurfaceShader(一)认识结构 转载请注明出处:http://www.cnblogs.com/-867259206/p/5595747.html 写 ...
- 如何用CSS进行开发
翻译自:How to Develop with CSS 很多web开发人员都知道CSS,但是他们很可能认为使用CSS的应用代码就是这样的: <p> <span style=" ...
- 翻译的技巧(The Technique of Translation)——前言
曰:你要发文章才能毕业! 答:谨遵圣命-- 缘起 研三了,要找工作了,要离开学校了,又要像浮萍般飘荡在世上了,还是学校好呀!本以为写个毕业论文就可以了,谁知老板幽幽地说写篇文章发了吧.我竟无言以对.不 ...
- 学习总结 JDBC
- IOS异步和多线程操作&&在sqlite3中的应用
1,数据库I/O操作(异步) 数据库本身是存储在磁盘上.访问和修改数据库,即对磁盘进行读写,即I/O操作. 磁盘属于计算机硬件,具有DMA能力,不需要CPU干预,可以实现异步操作. I/O操作一般是消 ...
- qq 登录 cordova插件
1.下载open-sdk.jar文件和mta_sdk_x.x.x.jar文件拷贝到libs(或lib)目录下这个链接 很容易的把环境配置好http://wiki.connect.qq.com/%E5% ...
- http://www.cnblogs.com/vowei/archive/2012/08/24/2654287.html
原创开源项目 - 扩展iQuery - 知平软件 - 博客园 return node.getProperty("mBottom").getValue();