HDU 2141

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 10874    Accepted Submission(s): 2846

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

题目很简单，就是给定N个（第一行输入的）不同的大数集，在接下来的N行里面分别输入他们的子集，要求判断你输入的某个数 X是否满足 Ai+Bj+Ck = X.这里的i,j,k表示下标不是乘法。在搜索符合题目要求的解时使用二分法，而不是暴力求解，节省编译时间。具体见代码。

#include
#include
#include
#include
using namespace std;
int L[501],M[501],N[501];
int Sum[250025];
int Count=1;
int main()
{
    int l,m,n,cnt;
    while(cin>>l>>m>>n)
    {
        int i,j,p=1;
        for(i=1;i<=l;i++)
            cin>>L[i];
        for(i=1;i<=m;i++)
            cin>>M[i];
        for(i=1;i<=n;i++)
            cin>>N[i];
        memset(Sum,0,sizeof(Sum));
        for(i=1;i<=l;i++)
            for(j=1;j<=m;j++)
                Sum[p++]=L[i]+M[j];
        sort(Sum,Sum+p-1);
        cout<<"Case "<>cnt;
        while(cnt--)
        {
            int s;
            cin>>s;
            int flag=0;  //用来记录sum的值是否符合要求
            for(i=1;i<=n;i++)
            {
                int start=1,ed=p-1;
                while(start<=ed)
                {
                    int mid=(start+ed)/2;
                    if(Sum[mid]+N[i]>s) {ed=mid-1;}
                    else if(Sum[mid]+N[i]