杭电2141Can you find it?

最新推荐文章于 2021-04-18 23:14:19 发布

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最新推荐文章于 2021-04-18 23:14:19 发布

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本文链接：https://blog.youkuaiyun.com/z8110/article/details/47152361

本文介绍了一种解决三数求和问题的有效算法。通过预处理两个数之和并使用二分查找法来快速判断是否存在第三个数使得三数之和等于给定的目标值。文章包含完整的C语言代码实现及样例输入输出。

#include <stdio.h>
#include <math.h>
#include <algorithm>
using namespace std;

int a1[505],a2[505],a3[505],y[255000],d[255000],v[255000];
int right,left,mid;
int i,j,k;
int a,b,c,t,s,u;
//int cmp(int u,int v)
//{
//	return u>v;
//}

int erfenfa(int x)
{
	left=0;right=u-1;
	while(left<=right)
	{
		mid=(left+right)/2;
		if(d[mid]<x)  left=mid+1;
		else if(d[mid]>x)right=mid-1;
		else if(d[mid]==x) 
		{
			k++;
			break;
		}
	}
	return 0;
}
int main()
{
    t=0;
	while(scanf("%d%d%d",&a,&b,&c)!=EOF)
	{
		t++;
		printf("Case %d:\n",t);
		for(int i=0;i<a;i++)
		  scanf("%d",&a1[i]);
		for(int i=0;i<b;i++)
		  scanf("%d",&a2[i]);
		for(int i=0;i<c;i++)
		  scanf("%d",&a3[i]);
		u=0;
		for(int i=0;i<b;i++)
		   for(int j=0;j<c;j++)
		    d[u++]=a2[i]+a3[j];
		sort(d,d+u);
		scanf("%d",&s);
		for(int i=0;i<s;i++)
		{
			scanf("%d",&y[i]);
			
			
			for(int j=0;j<a;j++)
			{
			v[j]=y[i]-a1[j];
		    k=0;
		    erfenfa(v[j]);
			if(k>0)
			   break;
		    }
			if(k>0)
				   printf("YES\n");
			else
				  printf("NO\n");
		}
	}
	return 0;
}

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
 
 
为什么要写这个题，个人感觉这道题挺有营养的，用的二分法，都说二分超时.第一次写的也超时了，但是看了看大神的代码，就把二分过程写到主函数外边，重新定义一个函数，发现a了，但是考虑之后感觉写到main函数里边跟重新定义函数没别的不同。莫非调用函数真的运行比较快吧；