【模拟】Sandglass

题目描述

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<r2<..<rK≤109
1≤Q≤105
0≤t1<t2<..<tQ≤109
0≤ai≤X(1≤i≤Q)
All input values are integers.

样例输入

180
3
60 120 180
3
30 90
61 1
180 180

样例输出

60
1
120



总之就是一个沙漏有A,B两半，X是A的容量，然后每个t时刻都会翻转一次。给你q个询问，每个询问给你a的初始值，问你t时刻a中有多少沙子

#include <bits/stdc++.h>
#define ll long long
using namespace std;
pair<int,int>team[100050];
int x,k,t[100050],q;
ll mn,mx,Now,ki,s;
int main()
{
    scanf("%d %d",&x,&k);
    for(int i=1;i<=k;i++)
    {
        scanf("%d",&t[i]);
    }
    scanf("%d",&q);
    for(int i=0;i<q;i++)
    {
        scanf("%d %d",&team[i].first,&team[i].second);
    }
    s=-1;
    mx=x;
    mn=0;
    ki=1;
    for(int i=0;i<q;i++)
    {
        while(ki<=k&&t[ki]<=team[i].first)
        {
            ll now=t[ki]-t[ki-1];
            now*=s;
            s*=-1;
            mn+=now;
            mx+=now;
            mn=max(mn,(ll)0);
            mn=min(mn,(ll)x);
            mx=max(mx,(ll)0);
            mx=min(mx,(ll)x);
            Now+=now;
            ki++;
        }
        ll now=team[i].first-t[ki-1];
        now*=s;
        ll ans=team[i].second+Now;
        if(ans<mn)
            ans=mn;
        if(ans>mx)
            ans=mx;
        ans+=now;
        ans=max(ans,(ll)0);
        ans=min(ans,(ll)x);
        printf("%lld
",ans);
    }
    return 0;
}