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
一开始TLE因为是三个for循环在一起，后来将两个for循环分开就没TLE了

#include<cstdio>#include<iostream>#include<algorithm>#include<cstdlib>int n,m,l,t,f,tmp,q;using namespace std;int a[1000],b[1000],c[1000],d[2555555];int main(){int cnt=0;    while(~scanf("%d%d%d",&l,&n,&m))    {f=0;    for(int i=0;i<l;i++)    scanf("%d",&a[i]);    for(int i=0;i<n;i++)    scanf("%d",&b[i]);    for(int i=0;i<m;i++)    scanf("%d",&c[i]);    int y=0;    for(int j=0;j<l;j++)    for(int k=0;k<n;k++)    d[y++]=a[j]+b[k];    sort(d,d+y);    scanf("%d",&t);    printf("Case %d:\n",++cnt);         while(t--)        {scanf("%d",&tmp);        for(int i=0;i<m;i++)         {q=tmp-c[i];        int low=0;            int high=y-1;            int mid=(y-1)/2;            while(low<high)                {                    if(d[mid]==q)                    {f=1;                    break;}                    if(d[mid]<q)                    low=mid+1;                    if(d[mid]>q)                    high=mid-1;                    mid=(high+low)/2;                }            if(d[low]==q || d[high]==q) f=1;            if(f)            break;}           if(f==1)            printf("YES\n");            else            printf("NO\n");            f=0;    }}    return 0;}