二分起步---Can you find it?

Can you find it?

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

这是一个基础的二分，只要将Ai+Bi=X-Ci,在Ai+Bi产生的结果里二分查找X-Ci就可以了。
AC代码：

#include<iostream>#include<algorithm>#include<cstring>using namespace std;int f[500*500+1],a[505],b[505],c[505];int len;void init(){    memset(a,0,sizeof(a));    memset(b,0,sizeof(b));    memset(c,0,sizeof(c));    memset(f,0,sizeof(f));}bool check(int k){    if(k==f[lower_bound(f,f+len,k)-f])        return true;    else        return false;}int main(){    int L,N,M;    int cas=1;    while(cin>>L>>N>>M)    {        init();        for(int i=0;i<L;i++)            cin>>a[i];        for(int i=0;i<N;i++)            cin>>b[i];        int k=0;        for(int i=0;i<L;i++)            for(int j=0;j<N;j++)                 f[k++]=a[i]+b[j];        sort(f,f+k);        len=k;        for(int i=0;i<M;i++)            cin>>c[i];        int T;        cin>>T;        cout<<"Case "<<cas++<<":"<<endl;        while(T--)        {            int x;            cin>>x;            int flag=0;            for(int i=0;i<M;i++)            {                if(check(x-c[i]))                {                    flag=1;                    break;                }            }            if(flag)                cout<<"YES"<<endl;            else                cout<<"NO"<<endl;        }    }    return 0;}

要开始突破二分了。
—–2017.11.11