hdu-2141（二分查找+暴力）

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
给A,B,C三个序列，分别从这三个序列中找三个数，若存在三个数之和等于X，则输出”YES”，否则输出”NO”

解析
三重循环的搜索肯定是不能用的。我的解法是将a+b的值存到ab数组中，将该数组排序。再用二分搜索判断x-c是否存在于该数组中，若存在，则证明存在x=a+b+c，输出’YES’即可。若对于所有c,都不存在x-c属于该数组,则证明不存在x=a+b+c，输出’NO’即可。存在一定的技巧，下面贴一波代码

代码

#include 
#include
#include

using namespace std;

int main()
{
    int a[510],b[510],c[510],ab[250010];
    int l,n,m;
    int s;
    int x[1010];
    int kase=0;
    bool ans;
    while(scanf("%d%d%d",&l,&n,&m)!=EOF){
        kase++;
        for(int i=0;i