HDU 2141 Can you find it?

题目：

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

题意：

有三个数组A、B、C，问是否存在Ai、Bi、Ci，使得Ai+Bi+Ci=X。

分析：

典型的二分题。如果用三个for循环暴力解决，时间复杂度为O(n^3)。考虑对其中两个数组求和并排序，再枚举将另一个数组，对和数组进行二分查找A+B=X-C。

#include <iostream>
#include <algorithm>
using namespace std;
const int max1=505;
int l,n,m,s,x,bci;   //bci即B、C数组合并后的元素数量
int a[max1],b[max1],c[max1*max1];
bool findx()
{

for(int i=0;i<l;++i)
{
int lo=0,hi=bci-1,mid;
while(lo<=hi)
{
mid=(lo+hi)>>1;
if(c[mid]+a[i]==x) return true;
else if(c[mid]+a[i]>x)  hi=mid-1;
else lo=mid+1;
}
}
return false;
}
int main()
{
int d=0;
while(cin>>l>>n>>m)
{
for(int i=0;i<l;++i)    cin>>a[i];
for(int i=0;i<n;++i)    cin>>b[i];
bci=0;
int temp;
for(int i=0;i<m;++i){
cin>>temp;
for(int j=0;j<n;++j){
c[bci++]=b[j]+temp;
}
}    //  c数组存放b+c
sort(c,c+bci);
cin>>s;     cout<<"Case "<<++d<<":"<<endl;
while(s--)
{
cin>>x;
//枚举a数组，二分c数组
if(findx())    cout<<"YES"<<endl;
else    cout<<"NO"<<endl;
}
}
return 0;
}

下面利用binary_search()，更简洁，但用时稍多。

#include <iostream>
#include <algorithm>
using namespace std;
const int max1=505;
int a[max1],b[max1],c[max1],bc[max1*max1];
int main()
{
int d=0;
int l,n,m,s,x,bci=0;
while(cin>>l>>n>>m)
{
bci=0;
for(int i=0;i<l;++i)    cin>>a[i];
for(int i=0;i<n;++i)    cin>>b[i];
for(int i=0;i<m;++i)    cin>>c[i];
for(int i=0;i<n;++i)
for(int j=0;j<m;++j)
bc[bci++]=b[i]+c[j];
sort(bc,bc+bci);
cin>>s;     cout<<"Case "<<++d<<":"<<endl;
while(s--)
{
cin>>x;
int cnt=0;
for(int i=0;i<l;++i)
{
if(binary_search(bc,bc+bci,x-a[i]))
{cnt=1;break;}
}
if(cnt)    cout<<"YES"<<endl;
else    cout<<"NO"<<endl;
}
}
return 0;
}