首先,先贴柳神的博客
想要刷好PTA,强烈推荐柳神的博客,和算法笔记
The ranklist of PAT is generated from the status list, which shows the scores of the submissions. This time you are supposed to generate the ranklist for PAT.
Each input file contains one test case. For each case, the first line contains 3 positive integers, N (≤104), the total number of users, K (≤5), the total number of problems, and M (≤105), the total number of submissions. 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 submission in the following format:
user_id problem_id partial_score_obtainedwhere partial_score_obtained is either ?1 if the submission 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.
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.
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 01 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 -submissions 意见书
corresponds to the full mark of the i-th problem
对应第i个问题的满分
either 两个中的任意一个
就是模拟PTA考试
具体的请百度
① 在赋值的时候,一定要考虑,原来的值,比如,一开始提交的时候是有分数的,后来提交编译就不通过了,这种情况下,编译不通过的那次是不可以覆盖点,原来的那次的 这是测试点4的问题之一
② 在算完美提交题目的时候,最好不要采用在线计算的方法,吧分数录入完成之后在算总分和完美提交题目的个数,因为,有的人可能会重复多次提交 -这也是测试点5 的问题
③ 关于代码的格式问题
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 -在每一行的最后 第一行的 18后面是不可以有空格的不然会报错
柳神的代码最后一行空了,没有事情,我的错误是别的错误,
最后一行也不可以有,不然测试点0会过不去
④ 不能输出的条件是全场都没有提交,或者是没有一个编译成功了,
也就是,只要有一个编译成功的,那我们就是要输出它
#include<iostream>
#include<algorithm>
#include<vector>
#include<string>
#include<cstring>
using namespace std;
struct node {
int problems[8] = {0};
int sum = 0,prefect=0,id;
bool flag=false; //判断该考生有没有
int rank;
};
bool cmp(node a,node b) {
if (a.sum != b.sum)
return a.sum > b.sum;
else if (a.prefect != b.prefect)
return a.prefect > b.prefect;
else
return a.id < b.id;
}
int main(void) {
int N, K, M;
scanf("%d%d%d", &N, &K, &M);
vector<node> data(N);
vector<int> prograde(K);
for (int i = 0; i < N; i++)
memset(data[i].problems, -1, sizeof(data[i].problems));
for (int i = 0; i < K; i++) {
scanf("%d", &prograde[i]);
}
for (int i = 0; i < M; i++) {
int id,problem, grade;
scanf("%d%d%d",&id, &problem, &grade);
if (grade >= 0) {
if (data[id-1].problems[problem] == -1) {
data[id - 1].problems[problem] = grade;
}
else if (data[id-1].problems[problem] != -1 && grade > data[id-1].problems[problem]) {
data[id-1].problems[problem] = grade;
}
data[id - 1].flag = true;
}
else {
if(data[id - 1].problems[problem]==-1)
data[id - 1].problems[problem] = 0;
}
}
for (int i = 0; i < N; i++) {
data[i].id = i + 1;
for (int j = 1; j <=K; j++) {
if (data[i].problems[j] >0) {
data[i].sum += data[i].problems[j];
}
if (data[i].problems[j] == prograde[j - 1]) {
data[i].prefect++;
}
}
}
sort(data.begin(), data.end(), cmp);
data[0].rank = 1;
for (int i = 1; i < N; i++) {
if (data[i].sum == data[i - 1].sum)
data[i].rank = data[i - 1].rank;
else
data[i].rank = i + 1;
}
bool first = false;
for (int i = 0; i < N; i++) {
if (data[i].flag) {
if (!first) {
first = true;
}
else {
printf("\n");
}
printf("%d %05d %d ", data[i].rank, data[i].id, data[i].sum);
for (int j = 1; j <= K; j++) {
if (data[i].problems[j] != -1) {
printf("%d", data[i].problems[j]);
if (j != K) {
printf(" ");
}
}
else {
printf("-");
if (j != K) {
printf(" ");
}
}
}
}
}
return 0;
} #include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
struct node {
int rank, id, total = 0;
vector<int> score;
int passnum = 0; //完美提交的题目数量
bool isshown = false; //要不要被输出
};
bool cmp1(node a, node b) {
if (a.total != b.total)
return a.total > b.total;
else if (a.passnum != b.passnum)
return a.passnum > b.passnum;
else
return a.id < b.id;
}
int main() {
int n, k, m, id, num, score;
scanf("%d %d %d", &n, &k, &m);
vector<node> v(n + 1);
for (int i = 1; i <= n; i++)
v[i].score.resize(k + 1, -1); //创建了K+1个元素,他们的值都是-1
vector<int> full(k + 1);
for (int i = 1; i <= k; i++)
scanf("%d", &full[i]);
for (int i = 0; i < m; i++) {
scanf("%d %d %d", &id, &num, &score);
v[id].id = id;
v[id].score[num] = max(v[id].score[num], score); //没有选用的,都是最好的score分数
if (score != -1)
v[id].isshown = true;
else if (v[id].score[num] == -1)
v[id].score[num] = -2;
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= k; j++) {
if (v[i].score[j] != -1 && v[i].score[j] != -2)
v[i].total += v[i].score[j];
if (v[i].score[j] == full[j])
v[i].passnum++;
}
}
sort(v.begin() + 1, v.end(), cmp1);
for (int i = 1; i <= n; i++) {
v[i].rank = i;
if (i != 1 && v[i].total == v[i - 1].total)
v[i].rank = v[i - 1].rank;
}
for (int i = 1; i <= n; i++) {
if (v[i].isshown == true) {
printf("%d %05d %d", v[i].rank, v[i].id, v[i].total);
for (int j = 1; j <= k; j++) {
if (v[i].score[j] != -1 && v[i].score[j] != -2)
printf(" %d", v[i].score[j]);
else if (v[i].score[j] == -1)
printf(" -");
else
printf(" 0");
}
printf("\n");
}
}
return 0;
}PTA甲级1075 PAT Judge (25point(s))
原文：https://www.cnblogs.com/a-small-Trainee/p/12455596.html