hdoj_2141

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 6083    Accepted Submission(s): 1585

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

3 3 3
1 2 3
1 2 3
1 2 3
3
1
4
10

 

Sample Output

Case 1:
NO
YES
NO

二分水题，开始没有预先保存a[i]+b[j]，wa。也算是个小小的优化吧

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

int main()
{
	freopen("in.txt","r",stdin);
	int a[505],b[505],c[505],sum[250050];
	int l,n,m,s,i,x,j;
	int ans = 0;
	int cnt = 1;
	int k = 0;
	while(scanf("%d %d %d",&l,&n,&m)!=EOF)
	{
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		memset(c,0,sizeof(c));
		memset(sum,0,sizeof(sum));
		for(i=0;i<l;i++)
			scanf("%d",&a[i]);
		for(i=0;i<n;i++)
			scanf("%d",&b[i]);
		for(i=0;i<m;i++)
			scanf("%d",&c[i]);
		k = 0;
		for(i=0;i<l;i++)
		{
			for(j=0;j<n;j++)
			{
				sum[k++] = a[i] + b[j];
			}
		}
		sort(sum,sum+k);
		sort(c,c+m);
		scanf("%d",&s);
		printf("Case %d:\n",cnt++);
		while(s--)
		{
			ans = 0;
			scanf("%d",&x);
			for(i=0;i<m;i++)
			{
				int left = 0;
				int right = l * n - 1;
				while(left<=right)
				{
					int mid = (left + right) / 2;
					if(sum[mid] + c[i] > x)
					{
						right = mid - 1;
					}
					else if(sum[mid] + c[i] < x)
					{
						left = mid + 1;
					}
					else 
					{
						printf("YES\n");
						ans = 1;
						break;
					}
				}
				if(ans==1) break;
			}
			if(ans==0) printf("NO\n");
		}
	}
	return 0;
}