hdu 2141 Can you find it? (二分法)

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 8506    Accepted Submission(s): 2216

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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 //375MS    1228K    1325 B    C++
 /*

     题意：
         给出三行数和Y，问每是否存在行中一个数令
            Ai+Bj+Ck==Y
         成立

     二分法：
         笨笨的分析错时间复杂度了
         先将两组数据合并，然后排序，然后进行二分
         时间复杂度：O(n*n*lgn)...

 */
 #include<stdio.h>
 #include<string.h>
 #include<stdlib.h>
 #define N 505
 int a[N],b[N],c[N],d[N*N];
 int l,n,m;
 int cmp(const void*a,const void*b)
 {
     return *(int*)a-*(int*)b;
 }
 int main(void)
 {
     int t,s;
     int cas=;
     while(scanf("%d%d%d",&l,&n,&m)!=EOF)
     {
         for(int i=;i<l;i++) scanf("%d",&a[i]);
         for(int i=;i<n;i++) scanf("%d",&b[i]);
         for(int i=;i<m;i++) scanf("%d",&c[i]);
         int cnt=;
         for(int i=;i<n;i++)
             for(int j=;j<m;j++)
                 d[cnt++]=b[i]+c[j];
         qsort(a,l,sizeof(a[]),cmp);
         qsort(d,cnt,sizeof(d[]),cmp);
         scanf("%d",&t);
         printf("Case %d:\n",cas++);
         while(t--){
             int flag=;
             scanf("%d",&s);
             for(int i=;i<l;i++){
                 int tl=,tr=cnt-;
                 int tt=s-a[i];
                 while(tl<tr){
                     int mid=(tl+tr)/;
                     if(d[tl]==tt || d[tr]==tt || d[mid]==tt){ //注意此处判断
                         flag=;break;
                     }else if(d[mid]<tt)
                         tl=mid+;
                     else tr=mid-;
                 }
                 if(flag) break;
             }
             if(flag) puts("YES");
             else puts("NO");
         }
     }
     return ;
 }