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 }