Can you find it?

最新推荐文章于 2023-04-16 15:29:27 发布

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本文介绍了一种解决三数求和问题的有效算法。通过将两个数序列组合并排序，利用二分查找来验证第三个数是否存在，使得问题得以高效解决。此方法适用于多个测试用例，确保了算法的准确性。

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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

Sample Output

Case 1:
NO
YES
NO

题解：二分题目，现将a，b合成一个数组，再将c移到右边，就相当于在新数组查找一个数，不过需要先排序，这是二分的特点。排序才nlogn。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int a[550];
int b[550];
int c[550];
int d[300000];

int main()
{
	int l,n,m;
	int cnt = 0;
	while(scanf("%d%d%d",&l,&n,&m) != EOF)
	{
		for(int i = 0;i < l;i++)
		{
			scanf("%d",a + i);
		}
		for(int i = 0;i < n;i++)
		{
			scanf("%d",b + i);
		}
		for(int i = 0;i < m;i++)
		{
			scanf("%d",c + i);
		}
		
		int k = 0;
		for(int i = 0;i < l;i++)
		{
			for(int j = 0;j < n;j++)
			{
				d[k++] = a[i] + b[j];
			}
		}
		
		sort(d,d + k);
		int num;
		int s;
		scanf("%d",&s);
		printf("Case %d:\n",++cnt);
		for(int i = 0;i < s;i++)
		{
			scanf("%d",&num);
			bool flag = false;
			for(int j = 0;j < m;j++)
			{
				int x = num - c[j];
				int y = lower_bound(d,d + k,x) - d;
				if(y != k && d[y] == x)
				{
					flag = true;
					break;
				}
			}
			if(flag)
			{
				printf("YES\n");
			}
			else
			{
				printf("NO\n");
			}
		}
	}
	
	return 0;
}