hdu-2141

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 29184    Accepted Submission(s): 7286

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

 

题意：给出三个数列ABC，从ABC各取一个整数求是否能够等于X，是则输出YES，否NO。

 

三组500直接查找会超时

可将A+B+C=X转换为A+B=X-C

之后即可将等式两边看做两个整体进行二分查找。

 

AC代码：

#include<bits/stdc++.h>
using namespace std;

int a[510],b[510],c[510];
int num[250010];

int bin(int q[],int x,int y){
    int left=0,right=x,mid;
    while(left<=right){
        mid=(left+right)/2;
        if(q[mid]==y)
        return 1;
        if(q[mid]>y)
        right=mid-1;
        else
        left=mid+1;
    }
    return 0;
}

int main(){
    int l,n,m,s,ans=0;
    while(cin>>l>>n>>m){
    ans++;
    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];
    }
    cin>>s;
    int x,cnt=0;
    
    for(int i=0;i<l;i++){
        for(int j=0;j<n;j++){
            num[cnt++]=a[i]+b[j];
        }
    }
    sort(num,num+cnt);
    printf("Case %d:\n",ans);
    while(s--){
        cin>>x;
        int flag=0;
        for(int i=0;i<m;i++){
            int temp=x-c[i];
            if(bin(num,cnt-1,temp)){
                cout<<"YES"<<endl;
                flag=1;
                break;
            }
        }
        if(!flag)
        cout<<"NO"<<endl;
    }
}
    return 0;
}