Codeforces-534D Handshakes

传送门

N个数，第i个数代表i之前的还未参加比赛的人的人数。每来够三个人可以参加比赛，她们可以随时加入比赛（也可以暂时不加入），求一个可能的到达顺序满足题意

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <vector>
 5 #define INF 0x3f3f3f3f
 6 #define MOD 1000000007
 7 using namespace std;
 8 typedef long long LL;
 9
10 const int maxn = 2e5 + 10;
11 int ans[maxn];
12 int cnt[maxn];
13 int num[maxn];
14 vector<int> G[maxn];
15
16 int nxt(int x) {
17     int up = x - 3, down = x % 3;
18     while (up > down) {
19         if (cnt[up]) return up;
20         up -= 3;
21     }
22     return down;
23 }
24
25 int main() {
26     int N;
27     scanf("%d", &N);
28     for (int i = 1; i <= N; i++) {
29         scanf("%d", &num[i]);
30         G[num[i]].push_back(i);
31         cnt[num[i]]++;
32     }
33     bool flag = 1;
34     int k = 0;
35     while (1) {
36         if (ans[0] == N) break;
37         if (cnt[k]) {
38             flag = 1;
39             int tmp = G[k][--cnt[k]];
40             ans[++ans[0]] = tmp;
41             k++;
42         } else {
43             k = nxt(k);
44             if (flag == 0) break;
45             flag = 0;
46         }
47     }
48     if (ans[0] == N) {
49         puts("Possible");
50         for (int i = 1; i <= ans[0]; i++) {
51             printf("%d%c", ans[i], i == ans[0] ? ‘\n‘ : ‘ ‘);
52         }
53     } else {
54         puts("Impossible");
55     }
56     return 0;
57 }