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

這個題目也提交了好多次

第一種方案是把2個數組的和用set來存，然後對每一個，在第3個數組裡面二分查詢。

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

int bs(int key,int a[],int length)
{
	int lo = 0, hi =length;
	int mi;
	while (lo <= hi)
	{
		mi = ((hi - lo) >> 1) + lo;//lo和hi比較大的時候相加可能爆int
		if (a[mi] == key)
 
 return mi;
		else if (a[mi]<key) lo = mi + 1;
		else hi = mi - 1;
	}
	return -1;//未找到
}


int main()
{
	int cas = 1;
	while (1)
	{
		int l, m, n;
		cin >> l >> m >> n;
		int *listl = new int[l];
		int *listm = new int[m];
		int *listn = new int[n];
		for (int i = 0; i < l; i++)cin >> listl[i];
		sort(listl, listl + l);
		for (int i = 0; i < m; i++)cin >> listm[i];
		for (int i = 0; i < n; i++)cin >> listn[i];
		set<int>se;
		for (int i = 0; i < m; i++)for (int j = 0; j < n; j++)

se.insert(listm[i] + listn[j]);
		set<int>::iteratorit;
		int s, sum;
		cin >> s;
		cout << "Case " << cas << ":" << endl;
		for (int i = 0; i < s; i++)
		{			
			cin >> sum;
			int temp = 0;
			for (it = se.begin(); it != se.end(); it++)
			{
				if (bs(sum - *it, listl,l-1) >= 0)
				{
					temp = 1;
					break;
				}
			}
			if (temp)cout << "YES";
			else cout << "NO";
			cout << endl;
		}
		cas++;
	}
	return 0;
}

但是結果超時了。

仔細一想，其實用set用處不大，因為25萬個int數，根本不會有太多重複的。

所以還是按照學長PPT裡面說的那種，2個數組求和，遍歷第三個陣列。

當然了，計重數就不用了。

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

int bs(int key,int a[],int length)
{
	int lo = 0, hi =length;
	int mi;
	while (lo <= hi)
	{
		mi = ((hi - lo) >> 1) + lo;//lo和hi比較大的時候相加可能爆int
		if (a[mi] == key) return mi;
		else if (a[mi]<key) lo = mi + 1;
		else hi = mi - 1;
	}
	return -1;//未找到
}

int main()
{
	int cas = 1;
	while (1)
	{
		int l, m, n;
		cin >> l >> m >> n;
		int *listl = new int[l];
		int *listm = new int[m];
		int *listn = new int[n];
		int *sum = new int[m*n];
		for (int i = 0; i < l; i++)cin >> listl[i];
		for (int i = 0; i < m; i++)cin >> listm[i];
		for (int i = 0; i < n; i++)
		{
			cin >> listn[i];
			for (int j = 0; j < m; j++)sum[i + j*n] = listn[i] + listm[j];
		}
		sort(sum, sum + m*n);
		int s, su;
		cin >> s;
		cout << "Case " << cas << ":" << endl;
		for (int i = 0; i < s; i++)
		{			
			cin >> su;
			int temp = 0;
			for (int i = 0; i < l;i++)
			{
				if (bs(su - listl[i], sum,m*n-1) >= 0)
				{
					temp = 1;
					break;
				}
			}
			if (temp)cout << "YES";
			else cout << "NO";
			cout << endl;
		}
		cas++;
		delete sum;
	}
	return 0;
}

不過，還是超時了。

這個時候，我開始感覺，可能是因為二分查詢的函式需要改進。

於是把二分查詢的函式變成了

int bs(int key, int a[], int length)
{
int lo = 0, hi = length;
if (key<a[0] || key>a[length])return -1;
int mi;
while (lo <= hi)
{
mi = ((hi - lo) >> 1) + lo;//lo和hi比較大的時候相加可能爆int
if (a[mi] == key) return mi;
else if (a[mi]<key) lo = mi + 1;
else hi = mi - 1;
}
return -1;//未找到
}
}

這個結果時候又變成了Output Limit Exceeded

實在找不出來哪裡輸出錯了，就百度了一下Output Limit Exceeded

剛好看到一個帖子http://bbs.csdn.net/topics/320153052

裡面說要儘量把while(1)這種的改成while(cin>>a)這種的

雖然以前在csuoj裡面用while(1)沒有出現這種問題，不過還是試了一下，結果真的AC了。