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
二分查询可以直接用set的find()方法,求和过大时要注意使用long long
AC代码:
#include<iostream>
#include<cstdio>
#include<set>
using namespace std;
long long al[510],an[510],am[510];
set<long long> subsum;
int main()
{
int caseNum=0;
int l,n,m;
while(cin>>l>>n>>m)
{
subsum.clear();
cout<<"Case "<<++caseNum<<":"<<endl;
for(int i=1;i<=l;i++)
scanf("%lld",&al[i]);
for(int i=1;i<=n;i++)
scanf("%lld",&an[i]);
for(int i=1;i<=m;i++)
scanf("%lld",&am[i]);
for(int i=1;i<=l;i++)
for(int j=1;j<=n;j++)
subsum.insert(al[i]+an[j]);
long long s,sum;cin>>s;
for(int x=1;x<=s;x++)
{
scanf("%lld",&sum);
int flag=0;
for(int i=1;i<=m;i++)
{
if(subsum.find(sum-am[i])!=subsum.end())
{
flag=1;break;
}
}
if(flag)cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
}
}