O - Can you find it?
Time Limit:3000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u
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
//很简单的题目
运用二分的思想去查找 a 和 b 组合的数组，是否有等于要求的
1 #include <iostream>
2 #include <algorithm>
3 #include <stdio.h>
4 using namespace std;
5
6 int a[505];
7 int b[505];
8 int c[505];
9 int sum[505*505];
10 int x,y,z,num;//数组a，b，c的元素个数
11
12 int find(int x)
13 {
14 int l=0,r=num-1;
15 int mid;
16 while (l<=r)
17 {
18 mid=(l+r)/2;
19 if (sum[mid]==x) return 1;
20 if (sum[mid]>x) r=mid-1;
21 if (sum[mid]<x) l=mid+1;
22 }
23 return 0;
24 }
25
26 int main()
27 {
28 int Case=0;
29 int i,j;
30 while (scanf("%d%d%d",&x,&y,&z)!=EOF)
31 {
32 for(i=0;i<x;i++)
33 scanf("%d",&a[i]);
34 for(i=0;i<y;i++)
35 scanf("%d",&b[i]);
36 for(i=0;i<z;i++)
37 scanf("%d",&c[i]);
38 sort(c,c+z);
39 num=0;
40 for (i=0;i<x;i++)
41 {
42 for (j=0;j<y;j++)
43 {
44 sum[num++]=a[i]+b[j];
45 }
46 }
47 sort(sum,sum+num);
48 int t,times;
49 scanf("%d",×);
50 printf("Case %d:
",++Case);
51 while (times--)
52 {
53 scanf("%d",&t);
54 for (i=0;i<z;i++)
55 {
56 if (find(t-c[i]))
57 {
58 printf("YES
");
59 break;
60 }
61 }
62 if (i==z)
63 printf("NO
");
64 }
65 }
66 return 0;
67 }