# 【二分】Can you find it?

Description
Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X.
Input
There are many cases. Every data case is described as followed: In the first line there are three integers L, N, M, in the second line there are L integers represent the sequence A, in the third line there are N integers represent the sequences B, in the forth line there are M integers represent the sequence C. In the fifth line there is an integer S represents there are S integers X to be calculated. 1<=L, N, M<=500, 1<=S<=1000. all the integers are 32-integers.
Output
For each case, firstly you have to print the case number as the form "Case d:", then for the S queries, you calculate if the formula can be satisfied or not. If satisfied, you print "YES", otherwise print "NO".
Sample Input
```3 3 3

1 2 3

1 2 3

1 2 3

3

1

4

10```
```Sample Output

Case 1:

NO

YES

NO```

```#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<math.h>
using namespace std;
int main()
{
long long l, m, n, cas = 1;
while (cin >> l >> m >> n)
{
long long a, b, c, d;
for (int i = 0; i < l; i++)
cin >> a[i];
for (int i = 0; i < m; i++)
cin >> b[i];
for (int i = 0; i < n; i++)
cin >> c[i];

int k = 0;
for (int i = 0; i < l; i++)
for (int j = 0; j < m; j++)
d[k++] = a[i] + b[j];
sort(d, d + k);
int s;
cin >> s;
printf("Case %d:\n", cas++);
while (s--)
{
long long x;
cin >> x;
bool flag = false;
for (int i = 0; i < n; i++)//第三个数组和新数组mid相加判断
{
int l = 0, r = k - 1;
while (l <= r)
{
int mid = (l + r) >> 1;
if (d[mid] + c[i] == x)
{
flag = true;
break;
}
else if (d[mid] + c[i] < x)
{
l = mid + 1;
}
else
r = mid - 1;
}
if (flag)
break;
}
if (flag)
cout << "YES" << endl;
else
cout << "NO" << endl;
}
}
return 0;
}```