标签：ams form int scan namespace amount printf bec lower
时间限制: 1 Sec 内存限制: 128 MB
原题链接 https://arc082.contest.atcoder.jp/tasks/arc082_d
We have a sandglass consisting of two bulbs, bulb A and bulb B. These bulbs contain some amount of sand. When we put the sandglass, either bulb A or B lies on top of the other and becomes the upper bulb. The other bulb becomes the lower bulb.
The sand drops from the upper bulb to the lower bulb at a rate of 1 gram per second. When the upper bulb no longer contains any sand, nothing happens.
Initially at time 0, bulb A is the upper bulb and contains a grams of sand; bulb B contains X?a grams of sand (for a total of X grams).
We will turn over the sandglass at time r1,r2,..,rK. Assume that this is an instantaneous action and takes no time. Here, time t refer to the time t seconds after time 0.
You are given Q queries. Each query is in the form of (ti,ai). For each query, assume that a=ai and find the amount of sand that would be contained in bulb A at time ti.
Constraints
1≤X≤109
1≤K≤105
1≤r1
180
3
60 120 180
3
30 90
61 1
180 180
60
1
120
我的做法是进行带优化的模拟，因为题目输入是确保后一个时间一定大于前一个时间的，所以对于翻转的模拟总共只需要按顺序进行一次便可，最后使用数学结论根据时间和初始质量推出答案。
需要特别注意的是，这道题会卡输入输出，如果不使用scanf/printf的话，必须要关闭同步锁且不能使用endl，不然会TLE。（可能是华东OJ的原因）
心疼那将近一半被输入输出卡掉的提交QAQ
42 ms/1664 KB（AtCoder）
80 ms/2080 KB（华东 OJ）
#include <iostream> #include <algorithm> using namespace std; int x, k, q, t, a, low, up, add, delta, now, ans, r[100007]; bool flag; int cal(int tt, int lower = 0, int upper = x) { if (tt < lower) return lower; if (tt > upper) return upper; return tt; } int main() { ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr); cin >> x >> k; for (int i = 1; i <= k; i++) cin >> r[i]; cin >> q; low = x, now = 1; while (q--) { cin >> t >> a; while (t >= r[now] && now <= k) { delta = (flag ? 1 : -1) * (r[now] - r[now - 1]), add += delta; up = cal(delta + up), low = cal(delta + low); flag = !flag, now++; } ans = cal((flag ? 1 : -1) * (t - r[now - 1]) + cal(a + add, up, low)); cout << ans << ‘\n‘; } return 0; }
标签：ams form int scan namespace amount printf bec lower
原文地址：https://www.cnblogs.com/xfl03/p/9385769.html
