K - Can you find it? HDU - 2141（二分法）

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.  

InputThere 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.  
OutputFor 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

暴力枚举，0（n^3)爆炸。

枚举前两个数，二分第三个O（n^2logn)

貌似也可能爆炸

先处理前两个数的和O（n^2)

排序，之后枚举第三个数

二分后两个数的和O(nlogn)

故复杂度为O(n^2)可以通过数据。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
    int l,n,m;
    int p=1;
    while(~scanf("%d%d%d",&l,&n,&m))
    {
        int a[510];
        int b[510];
        int c[510];
        long long ab[250005];
        for(int i=0; i<l; i++)scanf("%d",&a[i]);
        for(int i=0; i<n; i++)scanf("%d",&b[i]);
        for(int i=0; i<m; i++)scanf("%d",&c[i]);
        long long k=0;
        for(int i=0; i<l; i++)
        {
            for(int j=0; j<n; j++)
            {
                ab[k++]=a[i]+b[j];
            }
        }
        sort(ab,ab+k);

        printf("Case %d:\n",p++);
        int jj,j;
        scanf("%d",&jj);
        while(jj--)
        {
            scanf("%d",&j);
            int sign=0;
            for(int i=0; i<m; i++)
            {
                int xiao=0,da=k-1;
                int zhong=0;
                while(xiao<=da)//重点理解二分
                {
                    zhong=(xiao+da)/2;//直接从中间开始判断
                    if(ab[zhong]+c[i]==j)//如果其值正好与所求值相等
                    {
                        sign=1;//sign标记为1
                        break;//结束while循环
                    }
                    if(ab[zhong]+c[i]<j)//其值小于所求值
                        xiao=zhong+1;//小值赋值为“其值”加1
                    if(ab[zhong]+c[i]>j)//其值大于所求值
                        da=zhong-1;//大值赋值为“其值”减1
                }
                if(sign==1)//sign为1，表明有一个值等于所求值，结束for循环
                    break;
            }
            if(sign==1)
                printf("YES\n");
            else
                printf("NO\n");

        }
    }
    return 0;
}