CF1305E Kuroni and the Score Distribution(构造）

Posted 2021-10-22 goto_1600

Link
题意：

思路：
先构造1，n直到溢出了位置，然后考虑加n，如果单纯加n，那么对于前面贡献是$\frac{n}{2}$,然后发现n每隔2个数对前面的贡献都会减1，感性的理解，就是，之前n和一个值能匹配x，现在x变大了，n怎么匹配也匹配不了了，那么暴力加n知道刚好能满足m就行了，然后最妙的就是如何处理剩下的数，要造成0贡献，不如隔开构造，也就是说当前的最大值是max，前面有n个数，我们可以构造1e9-n*(maxv+1),1e9-(n-1)*(maxv+1)…
代码：

//#pragma GCC target("avx")
//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize("Ofast")
// created by myq
#include<iostream>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<climits>
#include<cmath>
#include<cctype>
#include<stack>
#include<queue>
#include<list>
#include<vector>
#include<set>
#include<map>
#include<sstream>
#include<unordered_map>
#include<unordered_set>
using namespace std;
typedef long long ll;
#define x first
#define y second
typedef pair<int,int> pii;
const int N = 400010;
const int mod=998244353;
inline int read()
{
	int res=0;
	int f=1;
	char c=getchar();
	while(c>'9' ||c<'0')
	{
		if(c=='-')	f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9')
	{
		res=(res<<3)+(res<<1)+c-'0';
	}
	return res;
 }
//#define int long long
	int n,m;
signed main()
{
	 int t;
//	 cin>>t;
	t=1;
	 int cas=0;
	 while(t--){
	 
	 	cin>>n>>m;
	 	int res=0;
	 	for(int i=1;i<=n;i++){
	 		res+=(i+1)/2-1;
		 }
//		 cout<<res<<endl;
		 if(res<m){
		 	cout<<"-1"<<endl;
		 }
		 else{
		 	int now=0;
		 	int maxv=0;
		 	bool flag=0;
		 	int tt;
			int id;
				int pos=0;
		 	for(int i=1;i<=n;i++)
			 {
		 		if(now+(i+1)/2-1<=m){
		 			cout<<i<<" ";
		 			now+=(i+1)/2-1;
				 }
				 else
				 {		
					 flag=1;	
					 id=i;			
//				 	cout<<now<<endl;
					tt=m-now;
					while(1){
						int cnt=0;
						for(int j=1;j<=i-1;j++)
							if(id-j<=i-1 && id-j>j)
							 cnt++;
//						cout<<endl;
//						cout<<id<<" "<<cnt<<endl;
//						break;
						if(cnt==tt)
						break;
						id++;
					}
					cout<<id<<" ";
					pos=i;
						 break;			
				 }
			 }
			 if(flag){
//			 	cout<<pos<<endl;
			 	for(int i=pos+1;i<=n;i++)
			 		cout<<(int)1e9-(n-i+1)*(id+1)<<" ";
			 }
			 cout<<endl;
		 }
	 }
	return 0;

}
/**
* In every life we have some trouble
* When you worry you make it double
* Don't worry,be happy.
**/