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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

10

4
1 2 3 4

#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;
}

HDOJ 5534 Virtual Participation 构造

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原文地址：http://blog.csdn.net/ck_boss/article/details/47255609