G - Traffic
vin is observing the cars at a crossroads. He finds that there are n cars running in the east-west direction with the i-th car passing the intersection at time ai . There are another m cars running in the north-south direction with the i-th car passing the intersection at time bi . If two cars passing the intersections at the same time, a traffic crash occurs. In order to achieve world peace and harmony, all the cars running in the north-south direction wait the same amount of integral time so that no two cars bump. You are asked the minimum waiting time.
input
The first line contains two integers n and m (1 ≤ n, m ≤ 1, 000). The second line contains n distinct integers ai (1 ≤ ai ≤ 1, 000). The third line contains m distinct integers bi (1 ≤ bi ≤ 1, 000).
output
Print a non-negative integer denoting the minimum waiting time.
Sample Input
1 1
1
1
1 2
2
1 3 Sample Output
1
0 题意:n辆车从一边经过十字路口,m辆车从另一边经过十字路口,当n方向的车经过路口时,m方向的车必须停下等待。给定车辆经过路口的时间,问m方向的车最少要等多久 题解:假设等待时间时t,当两个方向 的车同时经过路口时需要等,即b[j]+t==a[i],枚举输出最大的t即可
#include<iostream>
#include<string.h>
using namespace std;
int a[],b[];
int main()
{
int n,m,x;
while(cin>>n>>m)
{
memset(a,,sizeof(a));
memset(b,,sizeof(b));
for(int i=;i<n;i++)
{
cin>>x;
a[x]=;
}
for(int i=;i<m;i++)
cin>>b[i];
int t=;
for(int i=;i<m;i++)
{
if(a[b[i]+t])//如果b[i]+t==a[j]
{
t++;
i=-;
}
}
cout<<t<<endl;
}
return ;
}
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