HDOJ2141(Can you find it?)

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 888    Accepted Submission(s): 182


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
 
#include <iostream>
#include 
<algorithm>
using namespace std;

const int N = 502;

__int64 A[N],B[N],AB[N 
* N],C[N];
int ca, cb, cab, cc;

bool cmp(__int64 a, __int64 b)
{
    
return a < b;
}

void meso()
{
    
int i, j, k = 0;
    cab 
= ca * cb;
    
for(i = 0; i < ca; i++)
        
for(j = 0; j < cb; j++)
            AB[k
++= A[i] + B[j];
    sort(AB, AB 
+ cab, cmp);
}


bool search(__int64 TX)
{
    
int low = 0, high = cab - 1, mid;
    
while(low <= high)
    
{
        mid 
= (low + high) / 2;
        
if(TX == AB[mid])
            
return true;
        
else if(TX  < AB[mid])
            high 
= mid - 1;
        
else low = mid + 1;
    }

    
return false;
}


int main()
{
    
int i,S, kk = 1;
    
bool flag;
    __int64 X, TX;
    
while(cin>>ca>>cb>>cc)
    
{
        
for(i = 0; i < ca; i++)
            scanf(
"%I64d",A + i);
        
for(i = 0; i < cb; i++)
            scanf(
"%I64d",B + i);
        
for(i = 0; i < cc; i++)
            scanf(
"%I64d",C + i);
        meso();
        cout
<<"Case "<<kk++<<":"<<endl;
        cin
>>S;
        
while(S--)
        
{
            scanf(
"%I64d"&X);
            flag 
= false;
            
for(i = 0; i < cc; i++)
            
{
                TX 
= X - C[i];
                
if(search(TX))
                
{
                    flag 
= true;
                    
break;
                }

            }

            
if(flag)
                cout
<<"YES"<<endl;
            
else
                cout
<<"NO"<<endl;
        }

    }

    
return 0;
}

posted on 2009-04-11 19:32  Xredman  阅读(312)  评论(0编辑  收藏  举报

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