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。

 #include<cstdio>
 #include<cstdlib>
 #include<map>
 #include<set>
 #include<iostream>
 #include<cstring>
 #include<algorithm>
 #include<vector>
 #include<cmath>
 #include<stack>
 #include<queue>

 #define inf 0x7fffffff
 using namespace std;
 typedef long long LL;
 typedef pair<string, string> pr;

 int n, k, m;
 int sco[];
 const int maxn = 1e4 + ;
 struct node{
     bool print = false;
     int id;
     int s[] = {-, -, -, -, -, -, -};
     int sum = ;
     int num = ;
 }user[maxn];

 bool cmp(node a, node b)
 {
     if(a.sum == b.sum){
         if(a.num == b.num)return a.id < b.id;
         return a.num > b.num;
     }
     return a.sum > b.sum;
 }

 int main()
 {
     scanf("%d%d%d", &n, &k, &m);
     for(int i = ; i <= k; i++){
         scanf("%d", &sco[i]);
     }
     for(int i = ; i <= n; i++){
         user[i].id = i;
     }
     for(int i = ; i <= m; i++){
         string id;
         int pro, s;
         cin>>id>>pro>>s;
         int iid = stoi(id);
         if(s != -)user[iid].print = true;
         if(user[iid].s[pro] == -){
             if(s == -)user[iid].s[pro] = ;
             else user[iid].s[pro] = s;
             user[iid].sum += user[iid].s[pro];
             if(user[iid].s[pro] == sco[pro])user[iid].num++;
         }
         else if(user[iid].s[pro] < s){
             user[iid].sum += s - user[iid].s[pro];
             user[iid].s[pro] = s;
             if(user[iid].s[pro] == sco[pro])user[iid].num++;
         }
     }
     sort(user + , user +  + n, cmp);
     int rnk = ;
     for(int i = ; i <= n; i++){
         if(!user[i].print)continue;
         if(user[i].sum != user[i - ].sum)rnk = i;
         printf("%d %05d %d", rnk, user[i].id, user[i].sum);
         for(int j = ; j <= k; j++){
             if(user[i].s[j] != -)printf(" %d", user[i].s[j]);
             else printf(" -");
         }
         printf("\n");
     }

     return ;
 }