HDU - 2141 Can you find it?（二分）

Can you find it?

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

题目大意：在A、B、C三个数组中各选出一个数，看能否使得三个数之和为给定的数。只需讲其中两个数组中任意两个元素相加得到一个新的数组，再与剩下的数组中每一个元素进行二分即可

#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;

const int maxn = 510;

int a[maxn], b[maxn], c[maxn], d[maxn*maxn];
int cnt;

int bs(int x)
{
    int hi;
    int lo;
    int mid;
    lo = 0;
    hi = cnt-1;
    while(hi>=lo)
    {
        mid = (lo+hi) >> 1;
        if(d[mid]<x)
            lo = mid + 1;
        else if(d[mid]>x)
            hi = mid - 1;
        else
            return 1;
    }
    return 0;
}

int main()
{
    int l, m, n, k, s;
    int flag = 1;
    while(cin >> l >> m >> n)
    {
        for(int i = 0; i < l; i++)
        {
            cin >> a[i];
        }
        for(int i = 0; i < m; i++)
        {
            cin >> b[i];
        }
        for(int i = 0; i < n; i++)
        {
            cin >> c[i];
        }
        cnt = 0;
        for(int i = 0; i < m; i++)
        {
            for(int j = 0; j < n; j++)
            {
                d[cnt++] = b[i] + c[j];
            }
        }
        sort(d, d+cnt);
        sort(a, a+l);
        cin >> s;
        printf("Case %d:\n", flag++);
        int sum;
        while(s--)
        {
            cin >> sum;
            int ok = 0;
            /*if(a[0]+d[0]>sum || a[l-1]+d[cnt-1]<sum)
            {
                cout << "NO" << endl;
                continue;
            }*/
            //这段优化代码加进去就错了。。我也不知道为什么，有人如果能看出来原因请告诉我
            for(int j = 0; j < l; j++)
            {
                if(bs(sum-a[j]))
                {
                    ok = 1;
                    break;
                }
            }
            if(ok)
                cout << "YES" << endl;
            else
                cout << "NO" << endl;
        }
    }
    return 0;
}