HDU2141

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 
     
     
     
     

题目大意：给四组数，从第一二三组中分别找出一个，问是否能组成第四组中的每一个数。

解题思路：

从前两组中选出所有组合的可能，并存放在sum数组里，并对其排序，遍历第三组，在第三组中选个元素，在sum数组中二分查找是否存在与所选元素之和是第四组元素。

注意：

    STL 还是少用为好，占用空间大，运行效率差，若用set代替sum，的确是不用排序，但是会 MLE ，在别的题中也会出现用STL的运行时间是手写STL的两倍。。。

如果确保肯定能过，那就用，使代码简洁。

#include<iostream>
#include<stdlib.h>
#include<stdio.h>
#include<set>
#include<cstring>
#include<algorithm>

using namespace std;

int A[500];
int B[500];
int sum[250000];
int num, l, m, n;
int time = 0;

bool ans()
{
	for (int i = 0; i < n; ++i)
	{
		int l = 0, r = time-1;
		while (l < r)
		{
			int mid = (l + r) >> 1;
			if (sum[mid] < num - A[i])
			{
				l = mid + 1;
			}
			else
			{
				if (sum[mid] > num - A[i])
					r = mid - 1;
				else
					return true;
			}
		}
		if (sum[l] == num - A[i])
			return true;
	}
	return false;
}

int main()
{
	int coun = 0;
	while (scanf("%d%d%d", &l, &m, &n) != EOF)
	{
		time = 0;
		cout << "Case " << ++coun << ":" << endl;
		int i, j;
		for (i = 0; i < l; ++i)
			scanf("%d", &A[i]);
		for (i = 0; i < m; ++i)
			scanf("%d", &B[i]);
		for (i = 0; i < l; ++i)
			for (j = 0; j < m; ++j)
			{
				sum[time++] = A[i] + B[j];
			}
		sort(sum, sum + time);

		for (i = 0; i < n; ++i)
			scanf("%d", &A[i]);

		int s;
		scanf("%d", &s);
		for (i = 0; i<s; ++i)
		{
			scanf("%d", &num);
			if (ans())
			{
				cout << "YES" << endl;
			}
			else
			{
				cout << "NO" << endl;
			}
		}
	}
}