hdoj 2141 Can you find it? 【二分查找】

Can you find it?

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

Total Submission(s): 23825    Accepted Submission(s): 6039


[align=left]Problem Description[/align]
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.

 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
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".

 

[align=left]Sample Input[/align]

3 3 3
1 2 3
1 2 3
1 2 3
3
1
4
10

 

[align=left]Sample Output[/align]

Case 1:
NO
YES
NO

题目数据不算大，但是变量多啊，如果强制输入查找一定会暴，这题二分查找，对a[ ],b[ ]数组之和d[ ]查找可以省时间 
代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

int main()
{
int l,n,m,a[505],b[505],c[505],d[250005];
int s,x,cnt=0;
while(~scanf("%d%d%d",&l,&n,&m))
{
int ans=0;
for(int i=0;i<l;i++)
scanf("%d",&a[i]);
for(int j=0;j<n;j++)
{
scanf("%d",&b[j]);
for(int i=0;i<l;i++)
d[ans++]=a[i]+b[j];
}
for(int k=0;k<m;k++)
scanf("%d",&c[k]);
sort(d,d+ans);
scanf("%d",&s);
printf("Case %d:\n",++cnt);
for(int i=0;i<s;i++)
{
scanf("%d",&x);
int flag=0;
for(int k=0;k<m;k++)
{
int left=0,right=ans-1,mid;
while(left<=right)
{
mid=(left+right)/2;
if(c[k]+d[mid]==x)
{
flag=1;
break;
}
else if(c[k]+d[mid]<x)
left=mid+1;
else
right=mid-1;
}
if(flag)//没有这一步竟然超时了
break;
}
if(flag)
printf("YES\n");
else
printf("NO\n");
}
}
return 0;
}