HDU：2141 Can you find it?（二分+组合）

Can you find it?

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

Total Submission(s): 23766    Accepted Submission(s): 6019


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

 

Author

wangye

 

Source

HDU 2007-11 Programming Contest

 

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威士忌

题目大意：给你三个序列长度，下面跟上三个序列，接着跟上几次查询，问是否能在三个序列中分别找出一个元素，相加得到查询的值。

解题思路：先将前两个序列两两相加，得到一个新的序列，将新的序列排序，然后与第三个序列进行二分组合搜索。

代码如下：

#include <cstdio>
#include <algorithm>
using namespace std;
int l,n,m;
int a[510];
int b[510];
int c[510];
int niu[250010];//新a、b序列组成的新序列
int main()
{
int o=1;
while(scanf("%d%d%d",&l,&n,&m)!=EOF)
{
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]);
}
int cnt=0;//作为新数组的下标，同时计数新数组元素个数
for(int i=0;i<l;i++)//将前两个序列组合
{
for(int j=0;j<n;j++)
{
niu[cnt++]=a[i]+b[j];
}
}
sort(niu,niu+cnt);//排序，进行二分搜索必不可少的一步
int t;
scanf("%d",&t);
printf("Case %d:\n",o++);
while(t--)
{
int x;
scanf("%d",&x);
int flag=0;
int mid;
for(int i=0;i<m;i++)//将第三个序列的每个元素都与新序列进行二分搜索
{
int l=0,r=cnt-1;
while(l<=r)
{
mid=(l+r)/2;
if(niu[mid]+c[i]==x)
{
printf("YES\n");
flag=1;
break;
}
else
{
if(niu[mid]+c[i]>x)
{
r=mid-1;
}
else
{
l=mid+1;
}
}
}
if(flag==1)
break;
}
if(flag==0)
{
printf("NO\n");
}
}
}
}