问题 I: Sandglass
时间限制: 1 Sec 内存限制: 128 MB
提交: 238 解决: 41
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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
有一个沙漏，沙子总量为xx，初始情况上A下B，给出翻转时刻和多个询问，每个询问有询问时间和初始A中沙子量，求出询问时刻A中的沙子量。
#include <bits/stdc++.h>
#define N 100050
using namespace std;
pair<int,int>team[N];
int main()
{
int x;
int k;
int t[N]={0};
scanf("%d %d",&x,&k);
for(int i=1;i<=k;i++)scanf("%d",&t[i]);
int q;
scanf("%d",&q);
for(int i=0;i<q;i++)scanf("%d %d",&team[i].first,&team[i].second);
long long MIN=0,MAX=x,NOW=0,ki=1,sign=-1;
for(int i=0;i<q;i++)
{
while(ki<=k&&t[ki]<=team[i].first)
{
long long now=t[ki]-t[ki-1];
now*=sign;
sign*=-1;
MIN+=now;
MIN=max(MIN,(long long)0);
MIN=min(MIN,(long long)x);
MAX+=now;
MAX=max(MAX,(long long)0);
MAX=min(MAX,(long long)x);
NOW+=now;
ki++;
}
long long now=team[i].first-t[ki-1];
now*=sign;
long long ans=team[i].second+NOW;
if(ans<MIN)ans=MIN;
if(ans>MAX)ans=MAX;
ans+=now;
ans=max(ans,(long long)0);
ans=min(ans,(long long)x);
printf("%d\n",ans);
}
return 0;
}