PAT甲级1075 PAT Judge

题目：https://pintia.cn/problem-sets/994805342720868352/problems/994805393241260032

题意：

有m次OJ提交记录，总共有k道题，n个人。每道题有一个最高分。

现在要统计用户的排名，如果总分相同，完整AC的题目数高的排前面，都一样id小的排前面。

如果没有提交记录，或者提交记录都是-1的用户，就不输出。

思路：

根据题意模拟。PAT的题目都要耐心好好读题啊，各种情况都要看清楚。

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<map>
 4 #include<set>
 5 #include<iostream>
 6 #include<cstring>
 7 #include<algorithm>
 8 #include<vector>
 9 #include<cmath> 
10 #include<stack>
11 #include<queue>
12 
13 #define inf 0x7fffffff
14 using namespace std;
15 typedef long long LL;
16 typedef pair<string, string> pr;
17 
18 int n, k, m;
19 int sco[7];
20 const int maxn = 1e4 + 5;
21 struct node{
22     bool print = false;
23     int id;
24     int s[7] = {-1, -1, -1, -1, -1, -1, -1};
25     int sum = 0;
26     int num = 0;
27 }user[maxn];
28 
29 bool cmp(node a, node b)
30 {
31     if(a.sum == b.sum){
32         if(a.num == b.num)return a.id < b.id;
33         return a.num > b.num;
34     }
35     return a.sum > b.sum;
36 }
37 
38 int main()
39 {
40     scanf("%d%d%d", &n, &k, &m);
41     for(int i = 1; i <= k; i++){
42         scanf("%d", &sco[i]);
43     }
44     for(int i = 1; i <= n; i++){
45         user[i].id = i;
46     }
47     for(int i = 1; i <= m; i++){
48         string id;
49         int pro, s;
50         cin>>id>>pro>>s;
51         int iid = stoi(id);
52         if(s != -1)user[iid].print = true;
53         if(user[iid].s[pro] == -1){
54             if(s == -1)user[iid].s[pro] = 0;
55             else user[iid].s[pro] = s;
56             user[iid].sum += user[iid].s[pro];
57             if(user[iid].s[pro] == sco[pro])user[iid].num++;
58         }
59         else if(user[iid].s[pro] < s){
60             user[iid].sum += s - user[iid].s[pro];
61             user[iid].s[pro] = s;
62             if(user[iid].s[pro] == sco[pro])user[iid].num++;
63         }
64     }
65     sort(user + 1, user + 1 + n, cmp);
66     int rnk = 1;
67     for(int i = 1; i <= n; i++){
68         if(!user[i].print)continue;
69         if(user[i].sum != user[i - 1].sum)rnk = i;
70         printf("%d %05d %d", rnk, user[i].id, user[i].sum);
71         for(int j = 1; j <= k; j++){
72             if(user[i].s[j] != -1)printf(" %d", user[i].s[j]);
73             else printf(" -");
74         }
75         printf("\n");
76     }
77     
78     return 0;
79 }