hdu2141(二分)

最新推荐文章于 2019-05-05 12:09:59 发布

原创最新推荐文章于 2019-05-05 12:09:59 发布 · 338 阅读

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本文介绍了一道经典的三数求和算法题，通过预先计算两数之和并使用二分查找来高效解决该问题。输入包括三个整数序列及目标和，任务是判断是否存在三个数相加等于目标和。

Can you find it?

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 32334 Accepted Submission(s): 8061

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<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
typedef __int64 ll;
const int N=505;
ll a[N],b[N],c[N];
ll ab[N*N];
int l,n,m;	int cnt;
int BS(ll y)
{
	int left,right;
	left=0;right=cnt-1;
	int mid;
	while(left<=right)
	{
		mid=(left+right)/2;
		if(ab[mid]==y)
			return 1;
	    else if(ab[mid]<y)
	    {
	    	left=mid+1;
		}
		else right=mid-1;
	}
	return 0;
}
int main()
{
	int t=0;
	while(scanf("%d%d%d",&l,&n,&m)!=EOF)
	{
		t++;
		for(int i=0;i<l;i++)
		scanf("%I64d",&a[i]);
		for(int i=0;i<n;i++)
			scanf("%I64d",&b[i]);
		for(int i=0;i<m;i++)
			scanf("%I64d",&c[i]);
	    cnt=0;
		for(int i=0;i<l;i++)
			for(int j=0;j<n;j++)
			{
				ab[cnt++]=a[i]+b[j];
			}
		sort(ab,ab+cnt);
		sort(c,c+m);
		int q;
		ll x;
		scanf("%d",&q);	
		printf("Case %d:\n",t);
		while(q--)
		{
			scanf("%I64d",&x);
			int k;
			if(x<ab[0]+c[0]||x>ab[cnt-1]+c[m-1])
				puts("NO");
			else
			{
				ll g;
				for(k=0;k<m;k++)
				{
					g=x-c[k];
					if(BS(g))
					{
						puts("YES");
						break;
					}
				}
				if(k==m)
					puts("NO");
			}
			
		}
	}
	return 0;
}