HDU 2141 can you find it?【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): 21501    Accepted Submission(s): 5447

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

 

题目大意：给出三个集合A,B,C，给出一个数W,在集合A,B,C中各找到一个元素a，b，c，使得其加和等于W。

三层for时间复杂度较高，我们可以对其优化，我们两层for遍历A,B集合中元素，加和后放在一个数组det【】中。然后我们使用二分查找的方法，对w-c【i】进行查找，如果找到了w-c【i】，那么就输出YES,否则输出NO。

思考过程是在不断TLE中形成的T___T

AC代码：

#include<stdio.h>
#include<string.h>
#include<climits>
#include<algorithm>
using namespace std;
#define ll long long int
ll det[505*505];
ll a[505];
ll b[505];
ll c[505];
int cont;
int erfen(ll x)
{
    int l=0;int r=cont-1;
    while(r-l>=0)
    {
        int mid=(l+r)/2;
        //printf("%d %d %d\n",l,r,mid);
        if(det[mid]>x)
        {
            //printf("yes\n");
            r=mid-1;
        }
        else if(det[mid]<x)
        {
            //printf("yes2\n");
            l=mid+1;
        }
        else
        return 1;
    }
    return 0;
}
int main()
{
    int n1,n2,n3;
    int kase=0;
    while(~scanf("%d%d%d",&n1,&n2,&n3))
    {
        memset(det,0,sizeof(det));
        cont=0;
        for(int i=0;i<n1;i++)scanf("%I64d",&a[i]);
        for(int i=0;i<n2;i++)scanf("%I64d",&b[i]);
        for(int i=0;i<n3;i++)scanf("%I64d",&c[i]);
        for(int i=0;i<n1;i++)
        {
            for(int j=0;j<n2;j++)
            {
                det[cont++]=a[i]+b[j];
            }
        }
        sort(c,c+n3);
        sort(det,det+cont);
        int q;
        scanf("%d",&q);
        printf("Case %d:\n",++kase);
        while(q--)
        {
            ll w;
            scanf("%I64d",&w);
            int ok=0;
            for(int i=0;i<n3;i++)
            {
                if(erfen((ll)w-c[i])==1)
                {
                    ok=1;
                    printf("YES\n");
                    break;
                }
            }
            if(ok==0)printf("NO\n");
        }
    }
}