[HDU](2141)Can you find it? ---二分查找（查找）

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 31815    Accepted Submission(s): 7936

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 31 2 31 2 31 2 331410

 

Sample Output

Case 1:NOYESNO

 

解题新知： 
①题意：第一行给三个数l，n，m表示三个数组(A,B,C)的中元素的个数，下面三行一次是这三个数组的元素、
第四行给你一个数值t,表示要t次S。比如S=1，那么问这三个数组中是否存在Ai,Bj,Ck,使得Ai+Bj+Ck=1
若存在，那么输出"YES"，否则输出"NO"。
②思路：如果使用枚举那么复杂度为O（n^3）一定会超时，那么换一种思路，我们可以找任意两个数组，让他们构成一个新数组AB[]，即数组A和数组B的每一项互相组合相加，那么AB[l*n]，然后我么可以用二分查找的思想，tmp=S-c[i]，我们就可以在AB[I*n]中查找tmp，这样一转化，时间就得到了优化~

AC代码：

#include<iostream>#include<cstring>#include<algorithm>long long ab[250005];using namespace std;int bSearch(int first,int last,int key){    int mid;    while(first<=last)    {        mid=(first+last)/2;        if(ab[mid]==key)            return 1;        else if(ab[mid]<key)        {            first=mid+1;        }        else            last=mid-1;    }    return 0;}int main(){    int l,n,m;    int ccount=1;    while(cin>>l>>n>>m)    {        int a[505];        int b[505];        int c[505];        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        memset(c,0,sizeof(c));        memset(ab,0,sizeof(ab));        for(int i=0;i<l;i++)            cin>>a[i];        for(int i=0;i<n;i++)            cin>>b[i];        for(int i=0;i<m;i++)            cin>>c[i];        int sum=0;        for(int i=0;i<l;i++)        {            for(int j=0;j<n;j++)                ab[sum++]=a[i]+b[j];        }        sort(ab,ab+sum);        int t;        cin>>t;        cout<<"Case "<<ccount++<<":"<<endl;        while(t--)        {            int fn;            cin>>fn;            int tmp;            int i;            for(i=0;i<m;i++)            {                tmp=fn-c[i];                if(bSearch(0,sum-1,tmp))                {                    cout<<"YES"<<endl;                    break;                }            }            if(i==m)                cout<<"NO"<<endl;        }    }    return 0;}