问题 I: 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
思路：如果对每个问题一遍一遍的初始遍历，会超时。题目已经给出m个问题中，确保了时间是递增的。所以说我们不需要考虑初始状态是多少只需要将时间先行便利，之后把每一次的时间差值计算出来，但是这里需要一个判断，就是当沙子漏完时，还有时间就需要返回0，而不是负值当沙子大于初始值x时就需要返回最大值x。

#include<iostream>
#include<algorithm>
#include<stdio.h>
using namespace std;
int a[100010],x,n,m;
int judge(int xx,int yy=0,int w=x)//judge就是要判断每次当时间多 沙子不够时返回0 或者时间超出是返回最大值
{
	if(xx<yy) return yy;
	if(xx>w) return w;
	return xx;
}
int main()
{
	scanf("%d%d",&x,&n);
	a[0]=0;
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);

	scanf("%d",&m);

	int l=0,r=x,k=1;
	int sum=0,flag=-1;
	int t,b,cur;
	while(m--)
    {
		scanf("%d%d",&t,&b);

		while(t>=a[k]&&k<=n)
        {
			cur=flag*(a[k]-a[k-1]);//将每次需要漏下的差值计算出来 flag代表反转一次 加一次减一次
			sum+=cur;//再将这些数值相加
			l=judge(l+cur);
			r=judge(r+cur);
			k++;
			flag=-flag;
		}
		int ans;
		ans=judge(flag*(t-a[k-1])+judge(sum+b,l,r));//第一个judge判断离时间点最近的一个反转时间差值

		printf("%d\n",ans);
	}

	return 0;
 }