Can you find it? 二分

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 31 2 31 2 31 2 331410

Sample Output

Case 1:NOYESNO

前将前两组相加构成新数组，对新生成的进行二分
#include<iostream>#include<algorithm>using namespace std;int a[505],b[505],c[505],d[250005],l,n,m,x,s,tp,ed,mid,num=0;bool cmp(int A,int B){    return A<B;}int fun(int x){    for(int i=0;i<m;i++)    {        tp=0;        ed=l*n-1;        while(tp<=ed)        {            mid=(tp+ed)>>1;            if ((d[mid]+c[i])==x) return 1;            if ((d[mid]+c[i])<x) tp=mid+1;            else ed=mid-1;        }    }    return 0;}int main(){    while(cin>>l>>n>>m)    {        num++;        cout<<"Case "<<num<<":"<<endl;        for(int i=0;i<l;i++)            cin>>a[i];        for(int i=0;i<n;i++)            cin>>b[i];        for(int i=0;i<m;i++)            cin>>c[i];        for(int i=0;i<l;i++)            for(int j=0;j<n;j++)                d[i*n+j]=a[i]+b[j];        sort(d,d+l*n,cmp);        cin>>s;        while(s--)        {            cin>>x;            if (fun(x)) cout<<"YES"<<endl;            else cout<<"NO"<<endl;        }    }    return 0;}