CodeForces 767B The Queue

模拟。

情况有点多，需要仔细。另外感觉题目的$tf$有点不太对......而且数据水了。

$0$ $5$ $2$

$2$

$0$ $5$

这组数据按照题意的话答案可以是$2$和$4$，但是好多错的答案能$AC$。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<ctime>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0);
void File()
{
    freopen("D:\\in.txt","r",stdin);
    freopen("D:\\out.txt","w",stdout);
}
template <class T>
inline void read(T &x)
{
    char c = getchar();
    x = ;
    while(!isdigit(c)) c = getchar();
    while(isdigit(c))
    {
        x = x *  + c - '';
        c = getchar();
    }
}

LL INF=0x7FFFFFFF;
LL ts,tf,t;
int n;
LL s[];

int main()
{
    INF=INF*INF; INF=INF*;

    scanf("%lld%lld%lld",&ts,&tf,&t);
    scanf("%d",&n);
    for(int i=;i<=n;i++) scanf("%lld",&s[i]);

    if(n==)
    {
        printf("%lld\n",ts);
        return ;
    }

    if(s[]>ts)
    {
        printf("%lld\n",ts);
        return ;
    }

    LL need=INF,idx=-,last=ts,tmp;

    last=max(last,s[])+t;
    for(int i=;i<=n;i++)
    {
        if(s[i]==s[i-])
        {
            last=max(last,s[i])+t;
            continue;
        }
        if(s[i-]<=tf)
        {
            tmp=last+t;
            if(tmp<=tf&&tmp-s[i-]<need) idx=s[i-],need=tmp-s[i-];
        }
        if(s[i]-<=tf)
        {
            tmp=max(s[i]-,last)+t;
            if(tmp<=tf&&tmp-(s[i]-)<need) idx=s[i]-,need=tmp-(s[i]-);
        }
        if(last>=s[i-]&&last<s[i])
        {
            tmp=last+t;
            if(tmp<=tf&&tmp-last<need) idx=last,need=tmp-last;
        }

        last=max(last,s[i])+t;
    }

    if(s[]!=)
    {
        tmp=ts+t;
        if(<=tf)
        {
            if(tmp<=tf&&tmp-<need) idx=,need=tmp-;
        }
        tmp=max(s[]-,ts)+t;
        if(s[]-<=tf)
        {
            if(tmp<=tf&&tmp-(s[]-)<need) idx=s[]-,need=tmp-(s[]-);
        }
    }

    tmp=last+t;
    if(s[n]<=tf)
    {
        if(tmp<=tf&&tmp-s[n]<need) idx=s[n],need=tmp-s[n];
    }
    if(last<=tf)
    {
        if(tmp<=tf&&tmp-last<need) idx=last,need=tmp-last;
    }

    printf("%lld\n",idx);

    return ;
}