【二分/set/映射】 Can you find it? HDU基础04分治法

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最新推荐文章于 2024-08-13 10:26:25 发布

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本文介绍了一种解决三数求和问题的高效算法。通过预处理两个数之和并利用二分查找或哈希表进行快速匹配，该算法能够在多项式时间内找到三个数相加等于给定目标值的组合。

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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

Author

wangye

Source

HDU 2007-11 Programming Contest

#include <bits/stdc++.h>
using namespace std;
int l,n,m;
int a[505],b[505],c[505];
int d[250050];
int t=0;

void erfen(int X)
{
	for(int i=0;i<m;i++)  //枚举较小的数组
	{
		int x=X-c[i];

		int L=0,R=t-1;
		//搜索的数组范围

		while(L<=R)  
		{  //二分搜索较大的数组
			int mid=(L+R)/2;
			if(d[mid]<x)
				L=mid+1;
			else if(d[mid]>x)
				R=mid-1;
			else
			{
				printf("YES\n");
				return;
			}
		}
	}
	printf("NO\n");
	return;
}

int main()
{
	int ans=0;
	while(~scanf("%d%d%d",&l,&n,&m))
	{
		printf("Case %d:\n",++ans);
		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]);

		t=0;
		for(int i=0;i<l;i++)
			for(int j=0;j<n;j++)
			d[t++]=a[i]+b[j];
		sort(d,d+t);  //搜索前排序

		int s;
		scanf("%d",&s);
		while(s--)
		{
			int x;
			scanf("%d",&x);
			erfen(x);
		}
	}
	return 0;
}

#include <bits/stdc++.h>  //set 内存超限
using namespace std;
int l,n,m;
int a[505],b[505],c[505];
int d[250050];
set<int>ss;
int main()
{
    int ans=0;
    while(~scanf("%d%d%d",&l,&n,&m))
    {
        ss.clear();
        
        printf("Case %d:\n",++ans);
        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]);

        
        for(int i=0;i<l;i++)
            for(int j=0;j<n;j++)
            ss.insert(a[i]+b[j]);
        
        int s;
        scanf("%d",&s);
        while(s--)
        {
            int x;
            scanf("%d",&x);
            int flag=0;
            for(int i=0;i<m;i++)
            {
                if(ss.count(x-c[i]))
                {
                    flag=1;
                    break;
                }
            }
            if(flag) printf("YES\n");
            else printf("NO\n");
        }
    }
    return 0;
}

#include <bits/stdc++.h>
using namespace std;
const int mod=177777;  //玄学取模
int l,n,m;
int a[505],b[505],c[505];
int head[190050],tol;
struct node
{
    int to,next;
}rode[2500005];
bool find(int y)
{
    int x=y%mod;
    for(int i=head[x];i!=-1;i=rode[i].next)
        if(rode[i].to==y)return 1;
    return 0;
}
void add(int y)
{
    int x=y%mod;
    rode[tol].to=y;
    rode[tol].next=head[x];
    head[x]=tol++;
}
int main()
{
    int ans=0;
    while(~scanf("%d%d%d",&l,&n,&m))
    {
        tol=0;
        memset(head,-1,sizeof(head));
        printf("Case %d:\n",++ans);
        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]);


        for(int i=0;i<l;i++)
            for(int j=0;j<n;j++)
            if(!find((a[i]+b[j])))
            add(a[i]+b[j]);

        int s;
        scanf("%d",&s);
        while(s--)
        {
            int x;
            scanf("%d",&x);
            int flag=0;
            for(int i=0;i<m;i++)
            {
                if(find(x-c[i]))
                {
                    flag=1;
                    break;
                }
            }
            if(flag) printf("YES\n");
            else printf("NO\n");
        }
    }
    return 0;
}