ACM暑假集训_二分&三分_A-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

Hint

解题思路

有3个数表，问有多少种从3数表各选一个数的组合， 可以使得3个数的和加起来等于M。

直接每个组合方式都试一次然后判断时间复杂度太大，所以要有其他方法。

先将2，3个数表的全部数字相加，排序，然后再用二分法和第一个数表中的数字组合。

代码

#include <iostream>
#include<algorithm>
using namespace std;

int main()
{
	int L,N,M,S,X,m=1,s,min,max;
	int A[501],B[501],C[501],K[300000];
	while(cin>>L>>N>>M)
	{
		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];
		}
		int k=0;
		for(int i=0;i<N;i++)
		{ 
			for(int j=0;j<M;j++)
			{ 
				K[k++]=B[i]+C[j];
			} 
		} 
		sort(K,K+k);
		sort(A,A+L);
		cin>>S;
		cout<<"Case "<<m++<<":"<<endl;
		while(S--)
		{
			cin>>X;
			int flag=0;
			for(int i=0;i<L;i++)
			{ 
				int left=1,right=k-1;
				while(left<=right)
				{
					int mid=(right+left)/2;
					if(K[mid]+A[i]==X)
					{
						flag=1;
						break;
					}
					else if(K[mid]+A[i]<X)
					{
						left=mid+1;
					}
					else
					{
						right=mid-1;
					}
				}
			} 
			if(flag==1)
			{
				cout<<"YES"<<endl;
			}
			else
			{
				cout<<"NO"<<endl;
			}
		}
	}
	return 0;
}