Can you find it? --CSU-ACM2017暑假集训2-二分搜索

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本文介绍了一种解决特定三数求和问题的有效算法。该算法通过预先计算两个数的所有可能组合，并将这些组合存储在一个集合中以便快速查找，然后遍历第三个数并检查是否有匹配的组合来达到给定的目标值。此方法显著提高了搜索效率。

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

二分查询可以直接用set的find（）方法，求和过大时要注意使用long long

AC代码：

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

long long al[510],an[510],am[510];
set<long long> subsum;
int main()
{
    int caseNum=0;
    int l,n,m;
    while(cin>>l>>n>>m)
    {
        subsum.clear();
        cout<<"Case "<<++caseNum<<":"<<endl;
        for(int i=1;i<=l;i++)
            scanf("%lld",&al[i]);
        for(int i=1;i<=n;i++)
            scanf("%lld",&an[i]);
        for(int i=1;i<=m;i++)
            scanf("%lld",&am[i]);
        for(int i=1;i<=l;i++)
            for(int j=1;j<=n;j++)
                subsum.insert(al[i]+an[j]);
        long long s,sum;cin>>s;
        for(int x=1;x<=s;x++)
        {
            scanf("%lld",&sum);
            int flag=0;
            for(int i=1;i<=m;i++)
            {
                if(subsum.find(sum-am[i])!=subsum.end())
                {
                    flag=1;break;
                }
            }
            if(flag)cout<<"YES"<<endl;
            else cout<<"NO"<<endl;

        }

    }


}