Atocder ARC082 F-Sandglass 【思维题】*

Atocder ARC082 F-Sandglass

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Problem Statement

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

All input values are integers.

Input

The input is given from Standard Input in the following format:

Output

For each query, print the answer in its own line.

Sample Input 1

180
3
60 120 180
3
30 90
61 1
180 180

Sample Output 1

60
1
120
In the first query, 30 out of the initial 90 grams of sand will drop from bulb A, resulting in 60 grams. In the second query, the initial 1 gram of sand will drop from bulb A, and nothing will happen for the next 59 seconds. Then, we will turn over the sandglass, and 1 second after this, bulb A contains 1 gram of sand at the time in question.

Sample Input 2

100
1
100000
4
0 100
90 100
100 100
101 100

Sample Output 2

100
10
0
0
In every query, the upper bulb initially contains 100 grams, and the question in time comes before we turn over the sandglass.

Sample Input 3

100
5
48 141 231 314 425
7
0 19
50 98
143 30
231 55
342 0
365 100
600 10

Sample Output 3

19
52
91
10
58
42
100

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题目大意：有一个沙漏分成AB两面，里面共有X个点位的沙子，每一单位时间向下漏1个单位的沙子。
然后有n次操作，每次把沙漏翻转（不消耗时间）
询问当初始A有ai的沙子，B有X-ai的沙子，在t时刻A中有多少沙子

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考场上平方暴力

然后想想正解，挺神奇的一道题

首先我们可以用ai=0和X各按照规则跑一遍，跑出每个操作的A中沙子的上下界限

然后我们考虑预处理偏量，不考虑规则，预处理每个时刻的偏量

然后我们对于起始的量ai，直接加上偏量，如果在0和X处理出的区间内就合法，否则变成最靠近的一个

考虑如果直接处理偏量和0或X的折线相交会发生什么

首先明确ai只有可能在0或X触及界限的时候才会相交

那么在相交点右侧一定会存在一个转折点，那么在这个转折点位置ai一定比0低或比X高，所以在这之后ai一定不可能再回到0和X之间

然后就很简单了，是可以做到线性，但我懒，就多挂了一个log

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#include<bits/stdc++.h>
using namespace std;
#define N 100010
#define LL long long
int n,q;
LL X,r[N];
struct Node{LL a,t;}p[N];
LL l_line[N],r_line[N];
LL cnt[N];
int main(){
    scanf("%lld%d",&X,&n);
    r[0]=0;for(int i=1;i<=n;i++)scanf("%lld",&r[i]);
    scanf("%d",&q);
    for(int i=1;i<=q;i++)scanf("%lld%lld",&p[i].t,&p[i].a);
    LL na=0,nb=X;
    l_line[0]=0;r_line[0]=X;
    for(int tmp=1;tmp<=n;tmp++){
        if(tmp&1){
            LL tip=min(na,r[tmp]-r[tmp-1]);
            na-=tip;
            nb+=tip;
        }else{
            LL tip=min(nb,r[tmp]-r[tmp-1]);
            na+=tip;
            nb-=tip;
        }
        l_line[tmp]=na;
    }
    na=X,nb=0;
    for(int tmp=1;tmp<=n;tmp++){
        if(tmp&1){
            LL tip=min(na,r[tmp]-r[tmp-1]);
            na-=tip;
            nb+=tip;
        }else{
            LL tip=min(nb,r[tmp]-r[tmp-1]);
            na+=tip;
            nb-=tip;
        }
        r_line[tmp]=na;
    }
    for(int i=1;i<=n;i++)
        if(i&1)cnt[i]=cnt[i-1]-(r[i]-r[i-1]);
        else cnt[i]=cnt[i-1]+(r[i]-r[i-1]);
    for(int i=1;i<=q;i++){
        int ll=1,rr=n,res=0;
        while(ll<=rr){
            int mid=(ll+rr)>>1;
            if(r[mid]<=p[i].t)res=mid,ll=mid+1;
            else rr=mid-1;
        }
        LL pic=p[i].a+cnt[res];
        if(l_line[res]>pic)pic=l_line[res];
        if(r_line[res]<pic)pic=r_line[res];
        if(res&1)pic+=min(X-pic,p[i].t-r[res]);
        else pic-=min(pic,p[i].t-r[res]);
        printf("%lld
",pic);
    }
    return 0;
}