hdu 2141 Can you find it? 二分

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 9266    Accepted Submission(s): 2418

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

 

 

Recommend

威士忌   |   We have carefully selected several similar problems for you:   2899  2289  1551  2298  1399 

 

 1 /*

 2 题意：Ai+Bj+Ck = X;

 3            其中Ai 属于数组a[ ] , Bj 属于数组b[ ], Ck属于数组 C[ ] 的元素。

 4            求是否能找到满足X。而且X个数有1000，A,B,C个数为500.

 5            思路1.  枚举 X-Ck ，Bj，用二分找Ai是否存在。

 6                         时间复杂度 f(    1000*500*500*log(500)       )

 7            思路2.  Ai+Bj,看成一项。用数组保存tom[ 500*500+1];

 8                         枚举 X-Ck，用二分找tom[ ];

 9                         时间复杂度 f(     1000*500*log(500*500)        )

10 

11 */

12 

13 #include<iostream>

14 #include<stdio.h>

15 #include<cstring>

16 #include<cstdlib>

17 #include<algorithm>

18 using namespace std;

19 

20 int l[502],n[502],m[502];

21 int a[250003];

22 

23 int EF(int l,int r,int x)

24 {

25     int mid;

26     while(l<=r)  //  is  not  while(l<r)  !!

27     {

28         mid=(l+r)/2;

29         if( a[mid]>x)

30             r=mid-1;

31         else if( a[mid]<x)

32             l=mid+1;

33         else return 1;

34     }

35     return -1;

36 }

37 int main()

38 {

39     int i,j;

40     int L,N,M,S,num=0,x,k,len,t;

41     while(scanf("%d%d%d",&L,&N,&M)>0)

42     {

43         for(i=1;i<=L;i++)

44             scanf("%d",&l[i]);

45         for(i=1;i<=N;i++)

46             scanf("%d",&n[i]);

47         for(i=1;i<=M;i++)

48             scanf("%d",&m[i]);

49         len=L*N;t=0;

50         for(i=1;i<=L;i++)

51         for(j=1;j<=N;j++)

52         {

53             a[++t]=l[i]+n[j];

54         }

55         sort(a+1,a+1+len);

56         scanf("%d",&S);

57         printf("Case %d:\n",++num);

58         while(S--)

59         {

60             scanf("%d",&x);

61             for(i=1;i<=M;i++)

62             {

63                 k=EF(1,len,x-m[i]);

64                  if( k==1 ) break;

65             }

66             if( k==-1) printf("NO\n");

67             else printf("YES\n");

68         }

69     }

70     return 0;

71 }