A. Gennady the Dentist

 

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

 

This question must be clear order of things happen, first of all deal with v [i], followed by subsequent order of crying children, in fact, is directly next traversal, while traversing the side of the handle, stacking cry value, with a queue but self-defeating

 

Worried that p [i] will be too small, use a long, sample 56 that all children are crying, and the result p [i] burst into positive

 

 

 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     }

 

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Origin www.cnblogs.com/towerbird/p/11240514.html