Can you find it?
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
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
Source
HDU 2007-11 Programming Contest
/*
二分答案.
比较巧妙.
先将两个数组合并搞成一个n^2大的数组.
然后二分的话复杂度就有一个log.
二分和合并后的数组即对n^2取log.
然后复杂度就大大降低了.
*/
#include<iostream>
#include<algorithm>
#include<cstdio>
#define MAXN 501
#define LL long long
using namespace std;
LL s[MAXN*MAXN],a[MAXN],b[MAXN],c[MAXN],n1,n2,n3,n,m;
LL read()
{
LL x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9') x=x*10+ch-48,ch=getchar();
return x*f;
}
bool erfen(int l,int r,int i,int x)
{
int mid;
while(l<=r)
{
mid=(l+r)>>1;
if(s[mid]+c[i]==x) return true;
if(s[mid]+c[i]>x) r=mid-1;
else l=mid+1;
}
return false;
}
void slove()
{
int x;
bool flag;
while(m--)
{
flag=false;x=read();
for(int i=1;i<=n3;i++)
if(erfen(1,n,i,x)){flag=true;printf("YES
");break;}
if(!flag) printf("NO
");
}
return ;
}
int main()
{
int t=0;
while(~scanf("%d%d%d",&n1,&n2,&n3))
{
printf("Case %d:
",++t);n=0;
for(int i=1;i<=n1;i++) a[i]=read();
for(int i=1;i<=n2;i++) b[i]=read();
for(int i=1;i<=n3;i++) c[i]=read();
for(int i=1;i<=n1;i++)
for(int j=1;j<=n2;j++)
s[++n]=a[i]+b[j];
sort(s+1,s+n+1);
m=read();slove();
}
return 0;
}