6604: Sandglass
时间限制: 1 Sec 内存限制: 128 MB
提交: 347 解决: 79
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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
来源/分类
题意:给你一个沙漏，总容量是x，上面是a，下面是b，在k个时间点会将沙漏翻转，q次询问，每次在t的时间沙漏a初始为ai，最终沙漏a的沙子量。
题解:可以直接进行模拟处理，因为给定的时间一定是递增的，所以可以保证复杂度是k+q，每次求出两个翻转点之间沙子的变化量，根据变化量来限定能达到的最大上界，和能达到的最小下界，最终结果一定在这个上下界内。
#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6+7;
int x,q,r[maxn],k;
int cal(int tt,int lower=0,int upper=x)
{
if(tt<lower) return lower;
if(tt>upper) return upper;
return tt;
}
int main()
{
scanf("%d%d",&x,&k);
for(int i=1;i<=k;i++)
scanf("%d",&r[i]);
int q;
int delta=0,flag=0;
int now=1,low=x,up=0,a,add=0,ans,t;
scanf("%d",&q);
while(q--)
{
scanf("%d%d",&t,&a);
while(t>=r[now] && now<=k)
{
delta=(flag?1:-1)*(r[now]-r[now-1]),add+=delta;
low=cal(low+delta),up=cal(up+delta);
flag=!flag;now++;
}
ans=cal((flag?1:-1)*(t-r[now-1])+cal(a+add,up,low));
printf("%d\n",ans);
}
return 0;
}