HDU 2141 Can you find it?

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 3758    Accepted Submission(s): 935

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

 

 

这题是题好二分，要把两个数组合拼成一个。一个小地方错（代码里注明），让我找半天都找不到那里错。晕呀。。。现在还要搞清楚while(l < (=)r)的等号有与没的区别还有l = mid + 1和r = mid (- 1)要不要1。

还有如果数组从1...n和0...n-1的区别。要把这个弄清楚。

 

 

View Code

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 #define maxn 505
 4 
 5 int a[maxn], b[maxn], c[maxn], ab[maxn*maxn];
 6 int l, n, m;
 7 
 8 int Cmp(const void *a, const void *b)
 9 {
10     return *(int *)a - *(int *)b;
11 }
12 
13 int F(int n, int k)
14 {
15     int l, r, mid;
16     l = 0;
17     r = k - 1;//当时就是这里没多加个变量k,而直接l * m，搞得一直WA
18     while (l < r)
19     {
20         mid = (l + r) / 2;
21         if (ab[mid] == n)
22         {
23             return 1;
24         }
25         if (ab[mid] > n)
26         {
27             r = mid;
28         }
29         else
30         {
31             l = mid + 1;
32         }
33     }
34     return 0;
35 }
36 
37 
38 int main()
39 {
40     int   i, j, k, times = 1, flag, s, x;
41     
42     while (scanf("%d%d%d", &l, &m, &n) != EOF)
43     {
44         for (i = 0; i < l;scanf("%d", &a[i++]));
45         for (j = 0; j < m;scanf("%d", &b[j++]));
46         for (k = 0; k < n;scanf("%d", &c[k++]));
47         
48         
49         k = 0;
50         for (i = 0; i < l; i++)
51         {
52             for (j = 0; j < m; j++)
53             {
54                 ab[k++] = a[i] + b[j];
55             }
56         }
57         qsort(ab, k, sizeof(ab[0]), Cmp);
58 
59         printf("Case %d:\n", times++);
60         scanf("%d", &s);
61 
62         while (s--)
63         {
64             scanf("%d", &x);
65             flag = 1;
66             for (i = 0; i < n; i++)
67             {
68                 if (F(x - c[i], k) == 1)
69                 {
70                     flag = 0;
71                     break;
72                 }
73             }
74             if (flag == 0)
75             {
76                 puts("YES");
77             }
78             else
79             {
80                 puts("NO");
81             }
82         }
83     }
84     
85     return 0;
86 }