codeforce 767B

上周的cf比赛

B. The Queue
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Finally! Vasya have come of age and that means he can finally get a passport! To do it, he needs to visit the passport office, but it’s not that simple. There’s only one receptionist at the passport office and people can queue up long before it actually opens. Vasya wants to visit the passport office tomorrow.

He knows that the receptionist starts working after ts minutes have passed after midnight and closes after tf minutes have passed after midnight (so that (tf - 1) is the last minute when the receptionist is still working). The receptionist spends exactly t minutes on each person in the queue. If the receptionist would stop working within t minutes, he stops serving visitors (other than the one he already serves).

Vasya also knows that exactly n visitors would come tomorrow. For each visitor Vasya knows the point of time when he would come to the passport office. Each visitor queues up and doesn’t leave until he was served. If the receptionist is free when a visitor comes (in particular, if the previous visitor was just served and the queue is empty), the receptionist begins to serve the newcomer immediately.

“Reception 1”
For each visitor, the point of time when he would come to the passport office is positive. Vasya can come to the office at the time zero (that is, at midnight) if he needs so, but he can come to the office only at integer points of time. If Vasya arrives at the passport office at the same time with several other visitors, he yields to them and stand in the queue after the last of them.

Vasya wants to come at such point of time that he will be served by the receptionist, and he would spend the minimum possible time in the queue. Help him!

Input
The first line contains three integers: the point of time when the receptionist begins to work ts, the point of time when the receptionist stops working tf and the time the receptionist spends on each visitor t. The second line contains one integer n — the amount of visitors (0 ≤ n ≤ 100 000). The third line contains positive integers in non-decreasing order — the points of time when the visitors arrive to the passport office.

All times are set in minutes and do not exceed 1012; it is guaranteed that ts < tf. It is also guaranteed that Vasya can arrive at the passport office at such a point of time that he would be served by the receptionist.

Output
Print single non-negative integer — the point of time when Vasya should arrive at the passport office. If Vasya arrives at the passport office at the same time with several other visitors, he yields to them and queues up the last. If there are many answers, you can print any of them.

Examples
input
10 15 2
2
10 13
output
12
input
8 17 3
4
3 4 5 8
output
2
Note
In the first example the first visitor comes exactly at the point of time when the receptionist begins to work, and he is served for two minutes. At 12 minutes after the midnight the receptionist stops serving the first visitor, and if Vasya arrives at this moment, he will be served immediately, because the next visitor would only come at 13 minutes after midnight.

In the second example, Vasya has to come before anyone else to be served.

大意 ：就是一个人想去办理事情，想等待最少的时间去办理，现在他知道当天办公室的开门时间和工作人员对于每个人的接待时间（都是t）和他们的工作时间，然后他还知道当天去办理业务的每个人的到达时间，如果他和其他人在同时到达那里的话，他需要排在最后去办理，还有就是工作人员在剩余工作时间不足以完成一个人的业务办理的时候，他不会给新来的人办理业务。（他可以在零点到，办公室在零点后ts时开门，tf时关门，可以提前到，按到的先后排队）

这个题其实是贪心加模拟，我们可以把每个人遍历一遍，更新可以开始办的时间tim（tim=a【i-1】+t，即上一个人办完后），我们抢在当前这个人前面到即可（a【i】-1时到），注意tim+t不能超过tf，还有可以输出任意答案，只要正确。
然后还有一些很特殊的数据情况题上也没说，估计是hack出来的，我后来被坑了一次，我加在注释里。
代码

#include<iostream>#include<cstdio>#include<cstring>#define max(a,b)(a>b?a:b)#define min(a,b)(a<b?a:b)#define inf 2000000000000using namespace std;typedef long long ll;ll ts,tf,t,tim,ans;int n;ll a[100010];int main(){    //freopen("in.txt","r",stdin);    ll wat=inf;    scanf("%I64d%I64d%I64d%d",&ts,&tf,&t,&n);    for(int i=1;i<=n;++i){        scanf("%I64d",&a[i]);    }    tim=ts;    for(int i=1;i<=n;++i){        if(tim>tf-t)continue;        if(wat>tim-(a[i]-1)){        wat=tim-(a[i]-1);        ans=a[i]-1;        }        tim=max(tim,a[i])+t;    }       if(a[1]>tf)ans=ts;//坑爹的数据，第一个人比tf还晚，用上面的代码    //解决不了，直接在ts来就行，看完数据才明白用这有个情况。    if(tim<=tf-t)ans=tim;//如果最后一个人来的时间合适，也可以等人    //都办完再直接来，不用等。    printf("%I64d",ans);    return 0;}