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
题目分析：
按照顺序对翻转进行模拟，最后求出a中的沙子质量，注意使用scanf输入，不然会超时。
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
int ans,x,k,q,kk[100000+5],low,up,add,now,flag;
int ca(int r,int lower=0,int upp=x)
{
if(r<lower)
return lower;
if(r>upp)
return upp;
return r;
}
int main()
{
scanf("%d%d",&x,&k);
for(int i=1;i<=k;i++)
scanf("%d",&kk[i]);
now=1,flag=0,low=x;
scanf("%d",&q);
while(q--)
{
int t,a;
scanf("%d%d",&t,&a);
while(kk[now]<=t&&now<=k)
{
int dx;
dx=(flag?1:-1)*(kk[now]-kk[now-1]);
add+=dx;
low=ca(dx+low);
up=ca(dx+up);
now++;
flag=!flag;
}
ans =ca((flag ? 1 : -1) * (t - kk[now - 1]) + ca(a + add, up, low));
cout << ans << '\n';
}
return 0;
}