传送门：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.
题意：给a和b，分别是两天可以看书的时间，现在你有耗时1、2、3……的书，问你两天可以最多看几本书。
思路：要看的书尽量多，那么肯定是优先选择1、2、3……的书更优。那么找出刚好大于等于a+b的那个sum（1+到n），也就是说可以看n本书。因为你拥有的数字是从1~n，a+b=sum（1~n），也就是说a和b是可以刚好被这些数字耗完的，只要找出组成a或者b的数字，就可以求出另一天的情况了。那么怎么找出耗掉a的那些数字呢，一开始想的是从1开始，根据剩下的格子继续填，不行了就把前面的出队，后来想了想直接从最大的开始比较好，直接从n~1遍历填就能自动填完。
代码：
#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;
}