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Virtual Participation

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?

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

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.

10

4
1 2 3 4

假设答案只有A个1，B个2，C个3

则原题要找A+B+C<=Min(K,10^5) 且 A+B+C+AB+BC+AC=K 的解

右边方程移项得，(A+C+1)(B+C+1)=K+(1+C)^2-C

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (1000000+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
void work(ll k,ll &A,ll &B,ll &C)
{
	A=B=C=1;
	while (1)
	{
		int K2=k+(1+C)*(1+C)-C;
		
		for(ll x=(ll)(sqrt(K2));x>=C+1;x--) 
		{
			if (K2%x==0) {
				A=x-C-1,B=K2/x-C-1;  
				if (A+B+C>100000) continue;
				return; 
			}
			
		}
		C++;
	}

} 
int main()
{
//	freopen("H.in","r",stdin);
	
	ll k;
	
	while (cin>>k) {
		if (k==1) {
			puts("1");
			puts("1");
			continue;
		} else if (k==2) {
			puts("2");
			puts("1 1");
		} else if (k<=100000){
			printf("%d\n",k);
			For(i,k-1) printf("1 ");printf("1\n");
		} else {
			ll A,B,C=0;
			work(k,A,B,C);
			printf("%d\n",A+B+C);
			bool flag=0;
			For(i,A) 
			{
				if (!flag) flag=1;else printf(" ");
				printf("1");
			}
			For(i,B) 
			{
				if (!flag) flag=1;else printf(" ");
				printf("2");
			}
			For(i,C) 
			{
				if (!flag) flag=1;else printf(" ");
				printf("3");
			}printf("\n");
		}
		
		
		
		
		
	}
	
	return 0;
}

HDU 5334(Virtual Participation-(A+C+1)(B+C+1)=K+(1+C)^2-C)

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原文地址：http://blog.csdn.net/nike0good/article/details/47813717