HDU 2141（搜索题，二分）

最新推荐文章于 2019-04-10 14:41:49 发布

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#numbers #search #integer #ini #input

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本文探讨了一个经典的算法问题：给定三个整数序列A、B、C及目标整数X，判断是否存在A中的元素Ai、B中的元素Bj与C中的元素Ck，使得它们的和等于X。通过预处理和二分查找的方法提高了解决该问题的效率。

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Can you find it?

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 1836 Accepted Submission(s): 400

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

有人评论说是WA的，看来代码写的不够好，几年没接触算法的东西了。
#include <iostream>
#include <algorithm>
using namespace std;
#define N 505

__int64 ab[N*N];
int num;

int search(__int64 x)
{
	int f=0,l=num-1;
	int mid;
	while(f<=l)
	{
		mid=(f+l)/2;
		if(ab[mid]==x)
			return 1;
		else if(ab[mid]<x)
			f=mid+1;
		else
			l=mid-1;
	}
	return 0;
}

int main()
{
	int n,m,l,flag=0,s,i,j;
	__int64 a[N],b[N],c[N],x;
	while(scanf("%d%d%d",&n,&m,&l)!=EOF)
	{
		flag++;
		num=0;
		for(i=0;i<n;i++)
			scanf("%I64d",&a[i]);
		for(i=0;i<m;i++)
			scanf("%I64d",&b[i]);
		for(i=0;i<l;i++)
			scanf("%I64d",&c[i]);
		for(i=0;i<n;i++)
			for(j=0;j<m;j++)
				ab[num++]=a[i]+b[j];
		sort(ab,ab+num);
		sort(c,c+l);
		scanf("%d",&s);
		printf("Case %d:/n",flag);
		while(s--)
		{
			scanf("%I64d",&x);
			if(x<ab[0]+c[0] || x>ab[num-1]+c[l-1])
				printf("NO/n");
			else
			{
				int p;
				for(j=0;j<l;j++)
				{
					p=x-c[j];
					if(search(p))
					{
						printf("YES/n");
						break;
					}
				}
				if(j==l)
					printf("NO/n");
			}
		}
	}
	return 0;
}