The League of Sequence Designers(构造)

Posted 2021-12-23 Harris-H

The League of Sequence Designers(构造)

给定的算法就是连续最大子段和，而题意要求$maxlen\times sum$

考虑次优值的$sum=x$，最优值的$sum=x-1$，就是左边多个$-1$。
$$\begin{gathered} (x-1)\times len-x\times (len-1)=k \\ x-len=k \\ len<2000 \\ L\le len=x-k<2000 \\ x-k\le 1999 \\ x\le k+1999 \end{gathered}$$
因为要求$|a_i|\le 10^6$,因为要求$len\ge L$，不妨$len=1999$
这样$x=k+1999$

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull; 
const int N=1e3+5,M=2e4+5,inf=0x3f3f3f3f,mod=1e9+7;
const int hashmod[4] = 402653189,805306457,1610612741,998244353;
#define mst(a,b) memset(a,b,sizeof a)
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define x first
#define y second
#define pb emplace_back
#define SZ(a) (int)a.size()
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define ios ios::sync_with_stdio(false),cin.tie(nullptr) 
void Print(int *a,int n)
	for(int i=1;i<n;i++)
		printf("%d ",a[i]);
	printf("%d\\n",a[n]); 

template <typename T>		//x=max(x,y)  x=min(x,y)
void cmx(T &x,T y)
	if(x<y) x=y;

template <typename T>
void cmn(T &x,T y)
	if(x>y) x=y;

int main()
	int t,k,l;scanf("%d",&t);
	while(t--)
		scanf("%d%d",&k,&l);
		if(l>=2e3)
			puts("-1");continue;
		
		int n=2e3-1;
		printf("%d\\n-1",n);
		ll x=n+k;
		ll ave = x/(n-1),rest = x%(n-1);
		rep(i,2,n-1)
			printf(" %d",ave);	
		
		printf(" %d\\n",ave+rest);
	
	return 0;