hdu 2141 Can you find it? 二分

Can you find it?

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

Total Submission(s): 15781 Accepted Submission(s): 4034



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

二分，枚举其中一个，再枚举其他两个的和

#include <iostream>
#include <algorithm>
using namespace std;
int l,n,m;
long long a[555],b[555],c[555];
long long d[555*555];
int main()
{
int _case=0;
while(cin>>l)
{
cout<<"Case "<<++_case<<":"<<endl;
cin>>n>>m;
for(int i=1;i<=l;i++)
{
cin>>a[i];
}
for(int i=1;i<=n;i++)
{
cin>>b[i];
}
for(int i=1;i<=m;i++)
{
cin>>c[i];
}
int top=0;
for(int i=1;i<=l;i++)
{
for(int j=1;j<=n;j++)
{
d[++top]=a[i]+b[j];
}
}
sort(d+1,d+top+1);
sort(c+1,c+m+1);
int s;
cin>>s;
while(s--)
{
int ss;
cin>>ss;
int sumi=0,sumj=0;
if(d[1]+c[1]>ss||c[m]+d[top]<ss)
{
cout<<"NO"<<endl;
continue;
}
int flag=0;
for(int i=1;i<=m;i++)
{
int p=ss-c[i];
int l=1,r=top,mid=(r+l)/2;
while(l<=r)
{
if(d[mid]>p)
{
r=mid-1;
mid=(r+l)/2;
}
else if(d[mid]<p)
{
l=mid+1;
mid=(r+l)/2;
}
else
{
flag=1;
break;
}
}

}
if(flag) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
}
return 0;
}