A - Can you find it?解题报告

A - Can you find it?
Time Limit:3000MS Memory Limit:10000KB 64bit IO Format:%I64d
& %I64u
Submit Status

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

这个题目的意思呢就是从三组数中分别拉一个数出来，如果和==X的话就输出“YES”，否则输出“NO”；一开始虽然知道用硬的来时会超时，可是还是抱有侥幸的心态，结果TLE；还是用二分吧。这个二分首先将两个数组的和加起来，就是先开一个250001的数组，然后sort排序，然后呢就查找咯；

看代码把：

#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
int a[501],b[501],c[501],d[250001],L,N,M,k=0,i,j,qq=1,n,S;
while(cin>>L>>N>>M)
{
k=0;
for(i=0;i<L;i++)
cin>>a[i];
for(i=0;i<N;i++)
cin>>b[i];
for(i=0;i<M;i++)
cin>>c[i];
for(i=0;i<L;i++)
for(j=0;j<N;j++)
d[k++]=a[i]+b[j];
sort(d,d+k);
cin>>n;
cout<<"Case "<<qq<<':'<<endl;
for(i=0;i<n;i++)
{
cin>>S;
for(j=0;j<M;j++)
{
int temp=S-c[j];
int left=0,right=k-1,mid;
while(left<right)
{
mid=(left+right)/2;
if(d[mid]==temp)
{
cout<<"YES"<<endl;
goto end;
}
else if(temp>d[mid])
{
left=mid+1;
}
else
{
right=mid;
}
}
}
cout<<"NO"<<endl;
end:;
}
qq++;
}
return 0;
}