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