UPC6604 Sandglass（模拟）

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

#include <bits/stdc++.h>
#define N 100050
using namespace std;
//pair<int,int>s[N];
 struct node{
 int t,a;
 }s[N];
int main()
{
   int sand;
   int k;//k个时间点
   int time[N];
   scanf("%d %d",&sand,&k);
   for(int i=1;i<=k;i++)
    scanf("%d",&time[i]);
   int q;//q次查询
   scanf("%d",&q);
   for(int i=0;i<q;i++)scanf("%d %d",&s[i].t,&s[i].a);

   long long MIN=0,MAX=sand,NOW=0,ki=1,sign=-1;
   for(int i=0;i<q;i++)
   {
       while(ki<=k&&time[ki]<=s[i].t)//ki控制一共查询几次
       {
           long long now=time[ki]-time[ki-1];
           now*=sign;
           sign*=-1;

           MIN+=now;
           MIN=max(MIN,(long long)0);
           MIN=min(MIN,(long long)sand);
           MAX+=now;
           MAX=max(MAX,(long long)0);
           MAX=min(MAX,(long long)sand);

           NOW+=now;
           ki++;

       }

       long long now=s[i].t-time[ki-1];
       now*=sign;

       long long ans=s[i].a+NOW;
       if(ans<MIN)ans=MIN;
       if(ans>MAX)ans=MAX;

       ans+=now;

        ans=max(ans,(long long)0);
        ans=min(ans,(long long)sand);
       printf("%d\n",ans);
   }

    return 0;
}