Codeforces Round #398 (Div. 2) B. The Queue 思维

B. The Queue

链接：

http://codeforces.com/contest/767/problem/B

题解：

肯定要在某个人前一秒到达，到的更早没有意义，到的更晚就被他抢先了。 
这样可以O(n)枚举一遍解决。 
但是要考虑各种情况，比如在所有人之后到、别人到的时间都不在规定时间内。

代码：

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int, int> PII;
const ll mod = 1000000007;
const double eps = 1e-7;
// head

const int maxn = 1e5 + 7;
ll a[maxn];

int main()
{
    ll ts, tf, t, n;
    cin >> ts >> tf >> t >> n;
    rep(i, 1, n) scanf("%I64d", a + i);
    ll now = ts, ans, ansv = 1e15;
    if (a[1] > now) return 0 * printf("%I64d", now);

    for (int i = 1; i <= n; i++){
        if (max(0LL, now - a[i] + 1)<ansv && a[i] - 1 + t <= tf){
            ansv = max(0LL, now - a[i] + 1);
            ans = a[i] - 1;
        }
        now = max(now, a[i]) + t;
    }
    if (now + t <= tf) printf("%I64d\n", now);
    else printf("%I64d\n", ans);

    return 0;
}