构造

小于10^5的直接输出1

大于10^5的构造如下的串

1111...12222...233....3.....t,t+1,t+2..

设长度为n,每种字符出现了L[i]次则不同的串有  n*n+n-sigma(L[i])=2*K

大约估计一个n,用贪心求出L

Virtual Participation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Special Judge

Problem Description

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he asks rikka to have some practice on codeforces. Then she opens the problem B:

Given an integer K,
she needs to come up with an sequence of integers A satisfying
that the number of different continuous subsequence of A is
equal to k.

Two continuous subsequences a, b are
different if and only if one of the following conditions is satisfied:

1. The length of a is
not equal to the length of b.

2. There is at least one t that at≠bt,
where at means
the t-th
element of a and bt means
the t-th
element of b.

Unfortunately, it is too difficult for Rikka. Can you help her?

Input

There are at most 20 testcases，each testcase only contains a single integer K (1≤K≤109)

Output

For each testcase print two lines.

The first line contains one integers n (n≤min(K,105)).

The second line contains n space-separated
integer Ai (1≤Ai≤n) -
the sequence you find.

Sample Input

10

Sample Output

4
1 2 3 4

Author

XJZX

Source

2015 Multi-University Training Contest 4

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#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;

typedef long long int LL;

const int maxn=100100;

LL K,n;
LL L[maxn];

LL bs(LL x)
{
	LL low=1,high=100000;
	LL ans=1;
	while(low<=high)
	{
		LL mid=(low+high)/2;
		if(mid*(mid-1)<=x)
		{
			ans=mid;
			low=mid+1;
		}
		else
		{
			high=mid-1;
		}
	}
	return ans;
}

void solve(LL k)
{
    if(k<=100000)
    {
		cout<<k<<endl;
        for(int i=1;i<=k;i++)
            printf("%d%c",1,(i==k)?'\n':' ');
        return ;
    }
	int nt=0;
	n=1000;
	while(n*n+n-2*K<=0) n+=100;
	LL x=n*n+n-2*K;

	while(x)
	{
		LL y=bs(x);
		L[++nt]=y;
		x-=y*(y-1);
	}

	printf("%d\n",(int)n);
	int sum=0;
	int now=0;
	for(int i=1;i<=nt;i++)
	{
		sum+=L[i];
		for(int j=0;j<L[i];j++)
		{
			now++;
			printf("%d%c",i,(now==n)?'\n':' ');
		}
	}
	for(int i=0;i<n-sum;i++)
	{
		printf("%d%c",++nt,(now==n)?'\n':' ');
	}
	cout<<endl;
}

int main()
{
    while(cin>>K)
    {
        solve(K);
    }
    return 0;
}

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