hdu 2141 Can you find it?(二分)

最新推荐文章于 2020-03-25 15:16:25 发布

刘荣浩

最新推荐文章于 2020-03-25 15:16:25 发布

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分类专栏：二分/三分/尺取法文章标签：二分查找 hdu acm

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本文解析了一道经典的三数求和算法题，通过预处理两个数的所有可能和并进行排序，然后枚举第三个数并通过二分查找来判断是否存在满足特定条件的三个数。此方法适用于解决大规模数据集的问题。

Can you find it?

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 24218 Accepted Submission(s): 6127

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

题意：

给出一个等式Ai+Bj+Ck = X，并且给出A、B、C的取值，以及有S个X，查询在A、B、C中是否有值满足X关于等式成立.

思路：

我们需要枚举一个变量，那么，必须预处理出其中两个数的所有的和。这里，我预处理出A和B所有的可能，并且排序，枚举C，二分查找预处理出来的和。

代码：

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;
int L[510], N[510], M[510];
long long arr[250010];

int main(){
	int l, n, m, S, X, tot=0;
	while(cin>>l>>n>>m){
		for(int i=0; i<l; i++)
			cin>>L[i];
		for(int i=0; i<n; i++)
			cin>>N[i];
		for(int i=0; i<m; i++)
			cin>>M[i];

		int cnt = 0;
		for(int i=0; i<l; i++)
			for(int j=0; j<n; j++)
				arr[cnt++] = L[i]+N[j];

		sort(arr, arr+cnt);

		cout<<"Case "<<++tot<<":"<<endl;

		cin>>S;
		while(S--){
			cin>>X;
			int flag = -1;
			for(int i=0; i<m&&flag!=1; i++){
				int low = 0;
				int high = cnt-1;
				while(low<=high){
					int mid = (low+high)/2;
					if(M[i]+arr[mid]==X){
						flag = 1;
						break;
					}
					else if(M[i]+arr[mid]<X)
						low = mid+1;
					else
						high = mid-1;
				}
			}
			if(flag==1)
				cout<<"YES"<<endl;
			else
				cout<<"NO"<<endl;
		}

	}
	return 0;
}