HDU2141还是二分查找

Can you find it?

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others) Total Submission(s): 7898 Accepted Submission(s): 2052

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

Author

wangye

Source

HDU 2007-11 Programming Contest

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威士忌

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <cstdlib>
 6 #include <algorithm>
 7 #include <vector>
 8 #include <stack>
 9 #include <queue>
10 #include <cassert>
11 #include <set>
12 #include <sstream>
13 #include <map>
14 using namespace std ;
15 #ifdef DeBUG
16 #define bug assert
17 #else
18 #define bug //
19 #endif
20 #define zero {0}
21 const int MAXN=1000;
22 int L,M,N;
23 int a[MAXN],b[MAXN],c[MAXN];
24 int ab[300000];
25 int S,n;
26 int cmp(const void*a,const void*b)
27 {
28     return *(int *)a-*(int *)b;//升序 
29 }
30 int find(int min,int max,int x)
31 {
32     int mid;
33     while(min<=max)
34     {
35         mid=(min+max)/2;
36         if(ab[mid]==x)
37         return 1;
38         if(ab[mid]>x)
39         max=mid-1;
40         if(ab[mid]<x)
41         min=mid+1;
42     }
43     return 0;
44 }
45 int main()
46 {
47     #ifdef DeBUG
48         freopen("C:\Users\Sky\Desktop\1.in","r",stdin);
49     #endif
50     
51     int i,j,k;
52     int cas=1;
53     int num;
54     while(scanf("%d%d%d",&L,&N,&M)!=EOF)
55     {
56         num=0;
57         for(i=0;i<L;i++)
58         scanf("%d",&a[i]);
59         for(i=0;i<N;i++)
60         scanf("%d",&b[i]);
61         for(i=0;i<M;i++)
62         scanf("%d",&c[i]);
63         scanf("%d",&S);
64         for(i=0;i<L;i++)//将两个数组合并成一个数组，使其为两个数组中各个不同元素的和 
65         for(j=0;j<N;j++)
66         ab[num++]=a[i]+b[j];
67     //    qsort(c,M,sizeof(c[0]),cmp);//可排可不排 
68         qsort(ab,num,sizeof(ab[0]),cmp);//二分查找的前提是有序的数组 
69         printf("Case %d:
",cas++);
70         while(S--&&scanf("%d",&n))
71         {
72             bool _flag=false;
73             for(i=0;i<M;i++)
74             {
75                 if(find(0,num-1,n-c[i]))
76                 {
77                     _flag=true;
78                     break;
79                 }
80             }
81             if(_flag)
82             printf("YES
");
83             else
84             printf("NO
");
85         }
86     }
87     return 0;
88 }

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