Can you find it?

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 38634 Accepted Submission(s): 9395

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

Sample Output

Case 1:

YES

Author

wangye

Source

HDU 2007-11 Programming Contest

题目大意：

多组数据输入

第一行是三个序列各自元素的个数

第2，3，4是三个序列

第4行是x的个数

下面的x行是多种情况的x值

分析：

三个序列三个数组，把前面两个数组组合为一个数组

比如1 2 3，1 2 3，这两个序列组合为一个数组‘

1+1，1+2，1+3，2+1，2+2，2+3，3+1，3+2，3+3，

计算一下就是

2，3，4，3，4，5，4，5，6，有重复的但是不用去掉，称这个序列为t

然后与第三个序列组合

1 ，2，3，（第三个序列）

然后将t序列升序排序，t与第三个序列组合的时候，t序列采用二分寻找数据

注意点：

1.不能采用set去重，不然会超内存（第一次遇到超内存，好激动啊啊啊啊啊）

2.对t序列不能采用for循环寻找数据，要采用排好序之后再二分查找数据，不然会超时

代码如下：

#include<bits/stdc++.h>
using namespace std;
bool cmp(int a,int b)
{
    return a<b;
}
int main()
{
    int n1,n2,n3,n=1;
    while(~scanf("%d %d %d",&n1,&n2,&n3))
    {
        int a[n1],b[n2],c[n3];
        for(int i=0; i<n1; i++)
        {
            scanf("%d",&a[i]);
        }
        for(int j=0; j<n2; j++)
        {
            scanf("%d",&b[j]);
        }
        for(int k=0; k<n3; k++)
        {
            scanf("%d",&c[k]);
        }
        int xn;
        int t[n1*n2];
        scanf("%d",&xn);
        int x[xn];
        for(int i=0; i<xn; i++)
        {
            scanf("%d",&x[i]);
        }
        int y=0;
        for(int i=0; i<n1; i++)
        {
            for(int j=0; j<n2; j++)
            {
                int z=a[i]+b[j];
                t[y]=z;
                y++;
            }
        }
        sort(t,t+(n1*n2),cmp);
        printf("Case %d:
",n);
        for(int i=0; i<xn; i++)
        {
            int f=0;
            for(int k=0;k<n3;k++)
            {
                int l=0,h=n1*n2-1;
                while(l<=h)
                {
                    int mid=(l+h)/2;

                    if(t[mid]==x[i]-c[k])
                    {
                        f=1;
                        break;
                    }else if(t[mid]>x[i]-c[k])
                    {
                        h=mid-1;
                    }else if(t[mid]<x[i]-c[k])
                    {
                        l=mid+1;
                    }
                }
                if(f==1)
                    break;
            }
            if(f==1)
            {
                printf("YES
");
            }
            else
            {
                printf("NO
");
            }
        }
        n++;
    }
    return 0;
}

哈哈哈哈哈哈哈哈哈哈

今天真的是超级开心

不准吐槽。。。。。

。。。。。。。。。。