A. Gennady the Dentist

http://codeforces.com/contest/585/problem/A

这道题必须明确事情发生的顺序,首先处理v[i],其次是顺序往后的哭泣的小孩,其实是可以直接往后遍历,一边遍历一边处理,叠加cry值,用队列反而弄巧成拙了

担心p[i]会过小,要用long,样例56就是所有小孩都哭泣了,结果p[i]爆成正的

 

 

 1     static final int maxn = 4100;
 2     static int n;
 3     static int[] v = new int[maxn], d = new int[maxn];
 4     static long[] p = new long[maxn];
 5     static ArrayList<Integer> ans = new ArrayList<>();
 6 
 7     public static void main(String[] args) {
 8         IO io = new IO();
 9         n = io.nextInt();
10 
11         for (int i = 0; i < n; i++) {
12             v[i] = io.nextInt();
13             d[i] = io.nextInt();
14             p[i] = io.nextInt();
15         }
16 
17         for (int i = 0; i < n; i++)
18             if (p[i] >= 0) {
19                 ans.add(i);
20 
21                 long cry = 0;
22                 for (int j = i + 1; j < n; j++) {
23                     if (p[j] < 0) continue;
24                     if ((p[j] -= v[i] + cry) < 0) cry += d[j];
25                     if (v[i] > 0) v[i]--;
26                 }
27             }
28 
29         io.println(ans.size());
30         for (int a : ans) io.print((a + 1) + " ");
31         io.println("");
32     }
原文地址:https://www.cnblogs.com/towerbird/p/11240514.html