Can you find it?

最新推荐文章于 2023-04-16 15:29:27 发布

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最新推荐文章于 2023-04-16 15:29:27 发布

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本文链接：https://blog.youkuaiyun.com/qq_40510246/article/details/79857774

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本文介绍了一道经典的三数求和算法题，通过预处理和二分查找的方法优化了时间复杂度，实现了高效求解。文章详细展示了算法思路及实现代码。

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Can you find it?

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 37626 Accepted Submission(s): 9190

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 
 

Author

wangye

Source

HDU 2007-11 Programming Contest

PS：将两个数组两两相加，再与另一数组相加，其中，与另一数组的计算过程中，用二分不断缩小区间，达到减小时间复杂度的目的。、

代码：

#include<algorithm>
#include<iostream>
#include<string.h>
#include<stdio.h>
#include<math.h>
#include<vector>
#include<map>
typedef long long ll;
using namespace std;
const int N=1e6+10;
int main()
{
    int l,n,m,kk=1;
    int a[550],b[550],c[550],d[260000];
    while(cin>>l>>n>>m)
    {
        int k=0;
        for(int i=0; i<l; i++)
            cin>>a[i];
        for(int i=0; i<n; i++)
            cin>>b[i];
        for(int i=0; i<m; i++)
            cin>>c[i];
        for(int i=0; i<l; i++)
            for(int j=0; j<n; j++)
                d[k++]=a[i]+b[j];
        sort(d,d+k);
        int s,x;
        cin>>s;
        cout<<"Case "<<kk++<<":"<<endl;
        while(s--)
        {
        int o=0;
            cin>>x;
            for(int i=0; i<m; i++)
            {
                if(binary_search(d,d+k,x-c[i]))
                {
                    o=1;
                    break;
                }
            }
            if(o)
                cout<<"YES"<<endl;
            else
                cout<<"NO"<<endl;
        }
    }
}