hdu 2141 Can you find it?(二分+枚举)

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 14103    Accepted Submission(s): 3620

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

 

Author

wangye

 题目分析:先计算出b+c,然后枚举a,看有没有b+c==x-a,hash和二分都可以,hash会快一些

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#define MAX 507using namespace std;typedef long long LL;int l,n,m,s;LL x,a[MAX],b[MAX],c[MAX],p[MAX*MAX];LL f ( LL a , LL b  ){    return a+b;}bool search ( LL a , LL x ){    int left = 1 , right = n*m , mid;    while ( left != right )    {        mid = left + right >>1;        if ( f(a,p[mid] ) == x ) return true;        if ( f(a,p[mid]) < x ) left = mid+1;        else right = mid;    }    return false; }bool solve ( LL x ){    for ( int i = 1 ; i <= l ; i++ )    {        if ( f( a[i] , p[1] ) > x ) continue;        if ( f( a[i] , p[n*m] ) < x ) continue;        if ( search ( a[i] , x ) ) return true;    }    return false;}int main ( ){    int cc = 1;    while ( ~scanf ( "%d%d%d" , &l , &n , &m ) )    {        for ( int i = 1 ; i <= l ; i++ )            scanf ( "%I64d" , &a[i] );        for ( int i = 1 ; i <= n ; i++ )            scanf ( "%I64d" , &b[i] );        for ( int i = 1 ; i <= m ; i++ )            scanf ( "%I64d" , &c[i] );        for ( int i = 1 ; i <= n ; i++ )            for ( int j = 1 ; j <= m ; j++ )                p[(i-1)*m+j] = b[i]+c[j];        sort ( p+1 , p+n*m+1 );        scanf ( "%d" , &s );        printf ( "Case %d:\n" , cc++ );        while ( s-- )        {            scanf ( "%I64d" , &x );            if ( solve ( x ) ) puts ( "YES" );            else puts ( "NO" );        }    }}