HDU 2141.Can you find it?-二分

Can you find it?

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 42890 Accepted Submission(s): 10395

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

代码:

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int l[1005],n[1005],m[1005],c[250005];
 4 int ef(int l,int n,int m){
 5     int mid;
 6     while(l<=n){
 7         mid=(l+n)/2;
 8         if(c[mid]>m)
 9         n=mid-1;
10         else if(c[mid]<m)
11         l=mid+1;
12         else
13         return -1;
14     }
15     return 1;
16 }
17 int main(){
18     int a,b,s,x,p;
19     int i,j,h,num=0,len,ans;
20     while(~scanf("%d%d%d",&a,&b,&s)){
21         for(i=0;i<a;i++)
22             scanf("%d",&l[i]);
23         for(i=0;i<b;i++)
24             scanf("%d",&n[i]);
25         for(i=0;i<s;i++)
26             scanf("%d",&m[i]);
27             h=0;
28             for(i=0;i<a;i++){
29                 for(j=0;j<b;j++)
30                 c[h++]=l[i]+n[j];
31             }
32             len=a*b;
33             sort(c,c+len);
34             scanf("%d",&p);
35             printf("Case %d:
",++num);
36             while(p--){
37                 scanf("%d",&x);
38             for(i=0;i<s;i++){
39                 ans=ef(1,len,x-m[i]);
40                 if(ans==-1)
41                 break;
42             }
43             if(ans==1)
44             printf("NO
");
45             else
46             printf("YES
");
47         }
48     }
49     return 0;
50 }