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=1;
    while(scanf("%d%d%d",&l,&n,&m)!=EOF)
    {
        for(int i=0;i<l;i++) scanf("%d",&a[i]);
        for(int i=0;i<n;i++) scanf("%d",&b[i]);
        for(int i=0;i<m;i++) scanf("%d",&c[i]);
        int cnt=0;
        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++)
                d[cnt++]=b[i]+c[j];
        qsort(a,l,sizeof(a[0]),cmp);
        qsort(d,cnt,sizeof(d[0]),cmp);
        scanf("%d",&t);
        printf("Case %d:
",cas++);
        while(t--){
            int flag=0;
            scanf("%d",&s);
            for(int i=0;i<l;i++){
                int tl=0,tr=cnt-1;
                int tt=s-a[i];
                while(tl<tr){
                    int mid=(tl+tr)/2;
                    if(d[tl]==tt || d[tr]==tt || d[mid]==tt){ //注意此处判断 
                        flag=1;break;
                    }else if(d[mid]<tt)
                        tl=mid+1;
                    else tr=mid-1;
                }
                if(flag) break;
            }
            if(flag) puts("YES");
            else puts("NO");
        }
    }
    return 0;
}