Hdu Can you find it?(二分答案)

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\n");break;}
        if(!flag) printf("NO\n");
    }
    return ;
}
int main()
{
    int t=0;
    while(~scanf("%d%d%d",&n1,&n2,&n3))
    {
        printf("Case %d:\n",++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;
}