（binary_search） Can you find it hdu2141

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)

Total Submission(s): 35782    Accepted Submission(s): 8831

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

思路：

此处可以用暴力求解法以及二分查找法（会节省更多时间）。而二分查找是有对应的STL的。此处是可以用binary_search,是在有序数列中确定给定的元素是否存在。

#include <iostream>

#include <algorithm>

using namespace std;

int a[],b[],c[],sum[],s[];

int main()

{

    int l,n,m,i,j,s1,T=;

    while(scanf("%d%d%d",&l,&n,&m)!=EOF)

    {

        T++;

        for(i=;i<l;i++)

            scanf("%d",&a[i]);

        for(i=;i<n;i++)

            scanf("%d",&b[i]);

        for(i=;i<m;i++)

            scanf("%d",&c[i]);

        for(i=;i<l;i++)

        {

            for(j=;j<n;j++)

                sum[n*i+j]=a[i]+b[j];

        }

        sort(sum,sum+l*n);

        scanf("%d",&s1);

        for(i=;i<s1;i++)

            scanf("%d",&s[i]);

        printf("Case %d:\n",T);

        for(i=;i<s1;i++)

        {

            for(j=;j<m;)

            {

                if(binary_search(sum,sum+l*n,s[i]-c[j]))

                    break;

                j++;

            }

            if(j!=m)

                printf("YES\n");

            else

                printf("NO\n");

        }

    }

    return ;

}