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
分析：
用10^4打表 用10^2枚举
我的lower_bound一直挂 只能手写二分 不过还好
代码：
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 using namespace std;
6
7 const int maxn = 505;
8 int a[maxn], b[maxn], c[maxn], d[maxn * maxn];
9
10 int cnt;
11 bool check(int num) {
12 int low = 0; int high = cnt - 1;
13 while(low <= high) {
14 int mid = ( low + high) >> 1;
15 if(d[mid] >= num) {
16 high = mid - 1;
17 } else {
18 low = mid + 1;
19 }
20 }
21 if(d[high + 1] == num) return true;
22 return false;
23 }
24
25 int main() {
26 int l, n, m;
27 int t = 1;
28 while(EOF != scanf("%d %d %d",&l, &n, &m) ) {
29 for(int i = 0; i < l; i++) scanf("%d",&a[i]);
30 for(int i = 0; i < n; i++) scanf("%d",&b[i]);
31 for(int i = 0; i < m; i++) scanf("%d",&c[i]);
32 cnt = 0;
33 for(int i = 0; i < l; i++) {
34 for(int j = 0; j < n; j++) {
35 d[cnt++] = a[i] + b[j];
36 }
37 }
38 sort(d, d + cnt);
39 int s;
40 int num;
41 printf("Case %d:\n", t++);
42 scanf("%d",&s);
43 while(s--) {
44 scanf("%d",&num);
45 bool flag = false;
46 for(int i = 0; i < m; i++) {
47 if(check(num - c[i]) ) {
48 flag = true;
49 break;
50 }
51 }
52 if(flag) puts("YES");
53 else puts("NO");
54 }
55 }
56 }