HDU 2141 Can you find it?

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

 

题意：

给定三个数组a，b，c，每个数组有若干个数（<=500个），再给定一个数s要你判断是否存在s=a[i]+b[j]+c[k]，存在一组数就输出YES，一组都不存在就输出NO。

思路：

A+B = X - C

先把a数组和b数组中的数相加成一个ab[500×500]的数组，这样就相当于ab[i]+c[j]=s;再变形一下，ab[i]=c[j]+s，只要在ab数组里用二分查找看能否把c[j]+s找出来。

 

 1 #include<stdio.h>

 2 #include<string.h>

 3 #include<algorithm>

 4 #define N 505

 5 

 6 __int64 ab[N * N];

 7 int num;

 8 

 9 bool _search(__int64 x)

10 {

11     int i = 0, l = num - 1;

12     int mid;

13     while(i <= l){

14         mid = (i + l) / 2;

15         if(ab[mid] == x)

16             return true;

17         else if(ab[mid] < x)

18             i = mid + 1;

19         else

20             l = mid - 1;

21     }

22     return false;

23 }

24 

25 int main()

26 {

27     int n, m, l, kase = 0, s;

28     __int64 a[N], b[N], c[N], x;

29     while(scanf("%d%d%d", &n, &m, &l) == 3){

30         kase++; num = 0;

31         for(int i = 0; i < n; i++)

32             scanf("%I64d", &a[i]);

33         for(int i = 0; i < m; i++)

34             scanf("%I64d", &b[i]);

35         for(int i = 0; i < l; i++)

36             scanf("%I64d", &c[i]);

37 

38         for(int i = 0; i < n; i++)

39             for(int j = 0; j < m; j++)

40                 ab[num++] = a[i] + b[j];

41         std::sort(ab, ab+num);

42         std::sort(c, c+l);

43 

44         scanf("%d", &s);

45         printf("Case %d:\n", kase);

46         while(s--){

47             scanf("%I64d", &x);

48             if(x < ab[0] + c[0] || x > ab[num-1] + c[l-1])

49                 printf("NO\n");

50             else{

51                 int flag = 0;

52                 for(int i = 0; i < l; i++){

53                     __int64 p = x - c[i];

54                     if(_search(p)){

55                         printf("YES\n");

56                         flag = 1;

57                         break;

58                     }

59                 }

60                 if(!flag)

61                     printf("NO\n");

62             }

63         }

64     }

65     return 0;

66 }