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CodeForces 1072C Cram Time【思維題】

傳送門:http://codeforces.com/problemset/problem/1072/C

C. Cram Time

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

In a galaxy far, far away Lesha the student has just got to know that he has an exam in two days. As always, he hasn't attended any single class during the previous year, so he decided to spend the remaining time wisely.

Lesha knows that today he can study for at most aa hours, and he will have bb hours to study tomorrow. Note that it is possible that on his planet there are more hours in a day than on Earth. Lesha knows that the quality of his knowledge will only depend on the number of lecture notes he will read. He has access to an infinite number of notes that are enumerated with positive integers, but he knows that he can read the first note in one hour, the second note in two hours and so on. In other words, Lesha can read the note with number kk in kkhours. Lesha can read the notes in arbitrary order, however, he can't start reading a note in the first day and finish its reading in the second day.

Thus, the student has to fully read several lecture notes today, spending at most aa hours in total, and fully read several lecture notes tomorrow, spending at most bb hours in total. What is the maximum number of notes Lesha can read in the remaining time? Which notes should he read in the first day, and which — in the second?

Input

The only line of input contains two integers aa and bb (0≤a,b≤1090≤a,b≤109) — the number of hours Lesha has today and the number of hours Lesha has tomorrow.

Output

In the first line print a single integer nn (0≤n≤a0≤n≤a) — the number of lecture notes Lesha has to read in the first day. In the second line print nn distinct integers p1,p2,…,pnp1,p2,…,pn (1≤pi≤a1≤pi≤a), the sum of all pipi should not exceed aa.

In the third line print a single integer mm (0≤m≤b0≤m≤b) — the number of lecture notes Lesha has to read in the second day. In the fourth line print mm distinct integers q1,q2,…,qmq1,q2,…,qm (1≤qi≤b1≤qi≤b), the sum of all qiqi should not exceed bb.

All integers pipi and qiqi should be distinct. The sum n+mn+m should be largest possible.

Examples

input

Copy

3 3

output

Copy

1
3 
2
2 1 

input

Copy

9 12

output

Copy

2
3 6
4
1 2 4 5

Note

In the first example Lesha can read the third note in 33 hours in the first day, and the first and the second notes in one and two hours correspondingly in the second day, spending 33 hours as well. Note that Lesha can make it the other way round, reading the first and the second notes in the first day and the third note in the second day.

In the second example Lesha should read the third and the sixth notes in the first day, spending 99 hours in total. In the second day Lesha should read the first, second fourth and fifth notes, spending 1212 hours in total.

題意:給a和b,分別是兩天可以看書的時間,現在你有耗時1、2、3……的書,問你兩天可以最多看幾本書。

思路:要看的書儘量多,那麼肯定是優先選擇1、2、3……的書更優。那麼找出剛好大於等於a+b的那個sum(1+到n),也就是說可以看n本書。因為你擁有的數字是從1~n,a+b=sum(1~n),也就是說a和b是可以剛好被這些數字耗完的,只要找出組成a或者b的數字,就可以求出另一天的情況了。那麼怎麼找出耗掉a的那些數字呢,一開始想的是從1開始,根據剩下的格子繼續填,不行了就把前面的出隊,後來想了想直接從最大的開始比較好,直接從n~1遍歷填就能自動填完。

程式碼:

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<string>
#include<iostream>
#include<map>
#include<vector>
#include<set>
#include<queue>
using namespace std;
typedef long long ll;

ll a,b,co;
vector<ll>ans,ans2;

int main()
{
	std::ios::sync_with_stdio(false);
	cin>>a>>b;
	ll sum=a+b;
	ll pos=0;
	for(ll i=0;;i++)
	{
		ll now=(1+i)*i/2;
		ll nxt=(2+i)*(i+1)/2;
		if(now<=sum&&sum<nxt)
		{
			pos=i;
			break;
		}
	}
	for(ll i=pos;i>=1;i--)
	{
		if(a>=i)
		{
			ans.push_back(i);
			a-=i;
			co++;
		}
		else
		{
			ans2.push_back(i);
		}
	}
	cout<<co<<endl;
	for(ll i=0;i<ans.size();i++)
	{
		cout<<ans[i];
		if(i<ans.size()-1)
			cout<<" ";
	}
	cout<<endl;
	cout<<pos-co<<endl;
	for(int i=0;i<ans2.size();i++)
	{
		cout<<ans2[i];
		if(i<ans2.size()-1)
			cout<<" ";
	}
	cout<<endl;
	return 0;
}