A - Can you find it?（简单二分）

Can you find it?

Time limit 3000 ms Memory limit 10000 kB OS Windows

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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

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题意：

输入三个大小为N的数组A,B,C,然后给出X，问是否存在这样的i,j,k使得Ai+Bj+Ck=X。

解题思路：

求得前两个数组的和，遍历第三个数组，对前两个数组的和二分查找x-C[i],注意这道题不能用long long 不然会爆内存，int是足够存的。

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Code：

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

const int maxn=500+5;
int A[maxn],B[maxn],C[maxn];
int sum[maxn*maxn];
int L,N,M;

int bs(int x)
{
    int lo=0,hi=L*N-1,mid;
    while(lo<=hi)
    {
        mid=((hi-lo)>>1)+lo;
        if(sum[mid]==x)
            return mid;
        else if(sum[mid]<x)
            lo=mid+1;
        else
            hi=mid-1;
    }
    return -1;
}

int main()
{
    int ca=1;
    while(cin>>L>>N>>M)
    {
        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++)
                sum[i*L+j]=A[i]+B[j];
        sort(sum,sum+L*N);


        int s;

        cout<<"Case "<<ca++<<":"<<endl;
        cin>>s;
        while(s--)
        {
            int num;
            cin>>num;
            int flag=0;
            for(int i=0;i<M;i++)
            {
                if(bs(num-C[i])!=-1)
                {
                    flag=1;
                    break;
                }
            }
            if(flag)
                cout<<"YES"<<endl;
            else
                cout<<"NO"<<endl;
        }
    }
    return 0;
}