HDU2141还是二分查找

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others) Total Submission(s): 7898    Accepted Submission(s): 2052

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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威士忌

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <cassert>
#include <set>
#include <sstream>
#include <map>
using namespace std ;
#ifdef DeBUG
#define bug assert
#else
#define bug //
#endif
#define zero {0}
const int MAXN=1000;
int L,M,N;
int a[MAXN],b[MAXN],c[MAXN];
int ab[300000];
int S,n;
int cmp(const void*a,const void*b)
{
    return *(int *)a-*(int *)b;//升序 
}
int find(int min,int max,int x)
{
    int mid;
    while(min<=max)
    {
        mid=(min+max)/2;
        if(ab[mid]==x)
        return 1;
        if(ab[mid]>x)
        max=mid-1;
        if(ab[mid]<x)
        min=mid+1;
    }
    return 0;
}
int main()
{
    #ifdef DeBUG
        freopen("C:\Users\Sky\Desktop\1.in","r",stdin);
    #endif
    
    int i,j,k;
    int cas=1;
    int num;
    while(scanf("%d%d%d",&L,&N,&M)!=EOF)
    {
        num=0;
        for(i=0;i<L;i++)
        scanf("%d",&a[i]);
        for(i=0;i<N;i++)
        scanf("%d",&b[i]);
        for(i=0;i<M;i++)
        scanf("%d",&c[i]);
        scanf("%d",&S);
        for(i=0;i<L;i++)//将两个数组合并成一个数组，使其为两个数组中各个不同元素的和 
        for(j=0;j<N;j++)
        ab[num++]=a[i]+b[j];
    //    qsort(c,M,sizeof(c[0]),cmp);//可排可不排 
        qsort(ab,num,sizeof(ab[0]),cmp);//二分查找的前提是有序的数组 
        printf("Case %d:
",cas++);
        while(S--&&scanf("%d",&n))
        {
            bool _flag=false;
            for(i=0;i<M;i++)
            {
                if(find(0,num-1,n-c[i]))
                {
                    _flag=true;
                    break;
                }
            }
            if(_flag)
            printf("YES
");
            else
            printf("NO
");
        }
    }
    return 0;
}