题目传送门
1 /*
2 题意：有若干个点在一个区间内来回移动，1m/s。
3 水题：n^2的复杂度能解决，注意时间可能大于一个周期，要取模
4 */
5 #include <cstdio>
6 #include <algorithm>
7 #include <cstring>
8 #include <cmath>
9 using namespace std;
10
11 typedef long long ll;
12 const int MAXN = 1e3 + 10;
13 const int INF = 0x3f3f3f3f;
14 ll x[MAXN], y[MAXN];
15 struct Question
16 {
17 ll x, y, t;
18 }q[MAXN];
19 int ans[MAXN];
20
21 int main(void) //Gym 100553K Knockout Racing
22 {
23 // freopen ("K.in", "r", stdin);
24 freopen ("knockout.in", "r", stdin);
25 freopen ("knockout.out", "w", stdout);
26
27 int n, m;
28 while (scanf ("%d%d", &n, &m) == 2)
29 {
30 for (int i=1; i<=n; ++i) scanf ("%I64d%I64d", &x[i], &y[i]);
31 for (int i=1; i<=m; ++i) scanf ("%I64d%I64d%I64d", &q[i].x, &q[i].y, &q[i].t);
32 for (int i=1; i<=m; ++i)
33 {
34 int cnt = 0;
35 for (int j=1; j<=n; ++j)
36 {
37 ll d = q[i].t % ((y[j] - x[j]) * 2);
38 ll pos = x[j];
39 if (pos + d <= y[j]) pos += d;
40 else pos = y[j] - (d - (y[j] - x[j]));
41 if (q[i].x <= pos && pos <= q[i].y) cnt++;
42 }
43 ans[i] = cnt;
44 }
45
46 for (int i=1; i<=m; ++i) printf ("%d
", ans[i]);
47 }
48
49 return 0;
50 }