hdu 2143 数组合并 二分

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 22312    Accepted Submission(s): 5627

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

题目大意：给你三个数列，然后给你若干个sum,问你能否从每个数列中各取出一个数，使得它们的和为sum，如果可以，输出YES,

反之输出NO。

思路分析：很明显暴力做会超时，应该采用先合并数组然后二分查找，时间复杂度降到了nlogn，但是在实现的时候还是有几点要注意，

首先是输出格式，先 输出Case，然后再输入s，另外由于粗心大意，low和high的声明写到了for循环外面，wa到死，沧桑。

代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#define mem(a) memset(a,0,sizeof(a))
using namespace std;
const int inf=0xffffff;
const int maxn=500+50;
__int64 a[maxn],b[maxn],c[maxn];
__int64 f[maxn*maxn];
int kase=0;
int main()
{
    int L,N,M;
    __int64 s,sum;
    while(scanf("%d%d%d",&L,&N,&M)!=EOF)
    {
        int num=0;
        for(int i=0;i<L;i++)
            scanf("%I64d",&a[i]);
        for(int i=0;i<N;i++)
            scanf("%I64d",&b[i]);
        for(int i=0;i<M;i++)
            scanf("%I64d",&c[i]);
        for(int i=0;i<L;i++)
        {
            for(int j=0;j<N;j++)
            {
                f[num++]=a[i]+b[j];
            }
        }
        sort(f,f+num);
       printf("Case %d: ",++kase);
        scanf("%I64d",&s);
        while(s--)
        {
            scanf("%I64d",&sum);
            int flag=0;
        for(int i=0;i<M;i++)
        {
             int low=0,high=L*N-1;
            if(flag) break;
            while(low<=high)
            {
            int mid=(high+low)/2;
            if(f[mid]==sum-c[i])
            {
                flag=1;
                break;
            }
            else if(f[mid]<sum-c[i]) low=mid+1;
             else  high=mid-1;
            }
        }
         if(flag) cout<<"YES"<<endl;
         else cout<<"NO"<<endl;
        }
    }
    return 0;
}

tip:直接调用二分查找函数binary_search也可以解决