Can you find it?（hdu 2141 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): 19416    Accepted Submission(s): 4891


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
 
lower_bound提交就会出错，binary_search()也可以过
 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <algorithm>
 5 #include <set>
 6 using namespace std;
 7 int a[502],b[502],c[502],x;
 8 __int64 su[502*502];
 9 int l,n,m,s,u;
10 int d=1,sum;
11 bool solve(int x)
12 {
13     int i,k;
14     int ans;
15     for(i=0;i<m;i++)
16     {
17         ans=x-c[i];
18         int l=0,r=u-1,mid,e;
19         while(l<=r)
20         {
21             mid=(l+r)/2;
22             if(su[mid]==ans)    return 1;
23             else if(su[mid]>ans)    r=mid-1;
24             else l=mid+1;
25         }
26     }
27     return 0;
28 }
29 int main()
30 {
31     int i,j;
32     freopen("in.txt","r",stdin);
33     while(scanf("%d%d%d",&l,&n,&m)!=EOF)
34     {
35         for(i=0;i<l;i++)    scanf("%d",&a[i]);
36         for(i=0;i<n;i++)    scanf("%d",&b[i]);
37         for(i=0;i<m;i++)    scanf("%d",&c[i]);
38         scanf("%d",&s);
39         u=0;
40         for(i=0;i<l;i++)
41             for(j=0;j<n;j++)
42                 su[u++]=a[i]+b[j];
43         sort(su,su+u);
44         printf("Case %d:\n",d++);
45         for(i=0;i<s;i++)
46         {
47             scanf("%d",&x);
48             if(solve(x))    printf("YES\n");
49             else     printf("NO\n");
50         }
51     }
52 }

 

posted @ 2015-12-21 18:48  御心飞行  阅读(215)  评论(0)  编辑  收藏  举报