Can you find it?

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)

Problem 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

Author

wangye

Source

HDU 2007-11 Programming Contest

/*二分答案.比较巧妙.先将两个数组合并搞成一个n^2大的数组.然后二分的话复杂度就有一个log.二分和合并后的数组即对n^2取log.然后复杂度就大大降低了.*/#include
    
     #include
     
      #include
      
       #define MAXN 501#define LL long longusing namespace std;LL s[MAXN*MAXN],a[MAXN],b[MAXN],c[MAXN],n1,n2,n3,n,m;LL read(){    LL x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){
   if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9') x=x*10+ch-48,ch=getchar();    return x*f;}bool erfen(int l,int r,int i,int x){    int mid;    while(l<=r)    {        mid=(l+r)>>1;        if(s[mid]+c[i]==x) return true;        if(s[mid]+c[i]>x) r=mid-1;        else l=mid+1;    }    return false;}void slove(){    int x;    bool flag;    while(m--)    {        flag=false;x=read();        for(int i=1;i<=n3;i++)          if(erfen(1,n,i,x)){flag=true;printf("YES\n");break;}        if(!flag) printf("NO\n");    }    return ;}int main(){    int t=0;    while(~scanf("%d%d%d",&n1,&n2,&n3))    {        printf("Case %d:\n",++t);n=0;        for(int i=1;i<=n1;i++) a[i]=read();        for(int i=1;i<=n2;i++) b[i]=read();        for(int i=1;i<=n3;i++) c[i]=read();        for(int i=1;i<=n1;i++)          for(int j=1;j<=n2;j++)            s[++n]=a[i]+b[j];        sort(s+1,s+n+1);        m=read();slove();    }    return 0;}