【杭电2141】Can you find it?

最新推荐文章于 2021-04-18 23:14:19 发布

原创最新推荐文章于 2021-04-18 23:14:19 发布 · 245 阅读

0 ·

CC 4.0 BY-SA版权

编程语言同时被 3 个专栏收录

307 篇文章

订阅专栏

杭电acm~bestcoder

111 篇文章

订阅专栏

~~二分

15 篇文章

订阅专栏

本文介绍了一个典型的三数组求和问题，通过预处理两个数组的和并进行排序，然后利用二分查找来验证第三个数组元素与前两个数组组合是否能构成特定的目标值。这是一种高效解决问题的方法。

Can you find it?

Time Limit:3000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u

Submit Status

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 
      
      
       思路：先将前两组数据分别相加，得到一个新数组，对新数组排序，再二分与第三个数组相加，看是否有解
      
      
       1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60

       #include<cstdio>
#include<cstdlib>
#include<algorithm>
using namespace std;
const int M=510;
__int64 a[M],b[M],c[M];
__int64 num[M*M]; 
int main()
{
	int l,n,m,t,s,test=1;
	while(~scanf("%d%d%d",&l,&n,&m))
	{
		for(int i=1;i<=l;i++)
		scanf("%I64d",&a[i]);
		for(int i=1;i<=n;i++)
		scanf("%I64d",&b[i]);
		for(int i=1;i<=m;i++)
		scanf("%I64d",&c[i]);
		t=0;
		for(int i=1;i<=l;i++)
		{
			for(int j=1;j<=n;j++)
			{
			num[++t]=a[i]+b[j];//前两组数相加
			}
		}
		sort(num+1,num+t+1);
		__int64 x;
		scanf("%d",&s);
		printf("Case %d:\n",test++);
		while(s--)
		{   scanf("%I64d",&x);
			bool sine=false;	
			for(int i=1;i<=m;i++)
			{
				int k=1,r=t;
				while(k<=r)
				{
				int mid=(r+k)/2;
					if(num[mid]+c[i]==x)				   
					{
						sine=true;
					    break;
					}
					else if(num[mid]+c[i]<x)
					k=mid+1;
					else
					r=mid-1;											
				}
				if(sine)
				break;
			}
			if(sine)
			printf("YES\n");
			else
			printf("NO\n"); 
		}
	}
	return 0;
}