hdu-2141 can you find it(二分查找)

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 17976    Accepted Submission(s): 4538

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 31 2 31 2 31 2 331410

 

Sample Output

Case 1:NOYESNO

 

//hdu-2141 can you find it?(二分查找)//题目大意：//给你三组数，当输入一个数x时，从这三组数任意选三个数a[i],b[j],c[k];判断x是否等于a[i],b[j],c[k]三者之和；//若是，输出YES,否则，输出NO;//解题思路;// 定义三个数组a[],b[],c[]用来存放这三组数;主要思路是先将前两个数组的数相加合成一个数组d[];//然后用二分在d[]数组里查找x-c[k];若能找到一个数c[k]使得x-c[k]=d[mid]=a[i]+b[j];打印YES,否则，打印NO. #include<cstdio>#include<algorithm>using namespace std;int main(){int l,n,m,s,d[250010],x,flag;int a[510],b[510],c[510];             //定义三个数组，分别存放三组数； int i,j,ca=1,k,len,left,right,mid;while(scanf("%d%d%d",&l,&n,&m)!=EOF){len=0;for(i=0;i<l;i++) scanf("%d",&a[i]);for(i=0;i<n;i++) scanf("%d",&b[i]);for(i=0;i<m;i++) scanf("%d",&c[i]);for(i=0;i<l;i++)                //将前两个数组合为一个数组并排序；  for(j=0;j<n;j++)   d[len++]=a[i]+b[j];sort(d,d+len);scanf("%d",&s);printf("Case %d:\n",ca++);    while(s--)    {    k=0;    flag=0;    scanf("%d",&x);    for(i=0;i<m;i++)             //二分查找     {       if(k)     break;   left=0,right=len-1;       while(left<=right)       {      mid=(left+right)/2;       if(d[mid]>x-c[i])          right=mid-1;       else if(d[mid]<x-c[i])         left=mid+1;       else if(d[mid]==x-c[i])       {       k=1;       break;}   }}if(k) printf("YES\n");else printf("NO\n");}}  return 0;}