6604: Sandglass
时间限制: 1 Sec 内存限制: 128 MB
提交: 223 解决: 33
[提交] [状态] [讨论版] [命题人:admin]
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
【小结】
真特么讨厌这题，因为我菜。
【题意】
有个沙漏，有AB两端，里面有总沙量X，初始时A在上B在下。每一秒会从上向下流一个单位的沙子。
然后规定了k个时间点，每个时间点瞬间把沙漏AB颠倒。
然后q次询问，每次询问输入两个数：t，a，表示询问若初始0秒A端有沙量a，那么 t 秒的A端有多少沙量？
【分析】
只考虑A端，当A端在上时，沙量递减，A端在下时，沙量递增。
A端的沙量最少为0，最多为X，受这个限制，沙子流完时，若没有翻转，则停留。
设一个变量maxsand初始为X模拟过程。
设一个变量minsand初始为0模拟过程。
那么当初值为a时，每个时刻的沙量值会在minsand~maxsand之间
设一个变量val初始为0，模拟过程，但不受限制，也就是可以为负，可以>X。
对于输入的t，a，求出<=t时刻的那个翻转时的val值，若超出minsand~maxsand，则回归一下。
再把val加上a，若超出0~X，则回归一下。
即为答案。。。
【代码】
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAX=2e5+5;
const int INF=0x3f3f3f3f;
template<class T>bool gmax(T &a,T b){return a<b?a=b,1:0;}
template<class T>bool gmin(T &a,T b){return a>b?a=b,1:0;}
ll n,q,t,a,r[MAX],X;
int main()
{
scanf("%lld%lld",&X,&n);
for(int i=1;i<=n;i++)scanf("%lld",&r[i]);
scanf("%lld",&q);
ll minsand=0,maxsand=X,val=0,k=1,sign=-1;
while(q--)
{
scanf("%lld%lld",&t,&a);
while(k<=n&&r[k]<=t)
{
ll v=sign*(r[k]-r[k-1]);
minsand+=v;
maxsand+=v;
val+=v;
gmax(minsand,0ll);
gmin(minsand,(ll)X);
gmax(maxsand,0ll);
gmin(maxsand,(ll)X);
sign*=-1;
k++;
}
ll now=val+a;
gmax(now,minsand);
gmin(now,maxsand);
now+=sign*(t-r[k-1]);
gmax(now,0ll);
gmin(now,(ll)X);
printf("%lld\n",now);
}
}