1809C. Sum on Subarrays · znzryb

## 思路讲解

[ABC - 370 - E - Avoid K Partition](/2024/11/01/20241101-6613cb35) 

这种子序列的问题，无论是dp还是构造，抓手就是以下标i为结尾的子序列。

以这个结构单元来分析，不重不漏，一点一点的来。

## AC代码

[https://codeforces.com/problemset/submission/1809/309628591](https://codeforces.com/problemset/submission/1809/309628591)

```cpp
// Problem: C. Sum on Subarrays
// Contest: Codeforces - Educational Codeforces Round 145 (Rated for Div. 2)
// URL: https://codeforces.com/problemset/problem/1809/C
// Memory Limit: 512 MB
// Time Limit: 2000 ms
// by znzryb
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define FOR(i, a, b) for (long long i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (long long i = (a); i >= (b); --i)
#define all(x) x.begin(),x.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define fi first
#define se second
#define pb push_back
#define SZ(a) ((int) a.size())

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
typedef pair<DB,DB> pdd;
constexpr ll MAXN=static_cast<ll>(1e6)+10,INF=static_cast<ll>(5e18)+3;

ll N,K,T,A[MAXN];

inline void solve(){
	cin>>N>>K;
	ll res=0;
	for(int i=1;i<=N;++i){
		if(res==K){
			A[i]=-999;
		}else if(res+i>K){
			ll rem=K-res;
			A[i]=-((i-1-rem)*2+1);
			res=K;
		}else{
			res+=i;
			A[i]=2;
		}
	}
	for(int i=1;i<=N;++i){
		cout<<A[i]<<" ";
	}
	cout<<"\n";
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	cin>>T;
	while(T--){
		solve();
	}
	return 0;
}
/*
AC
https://codeforces.com/problemset/submission/1809/309628591
*/
```

## 心路历程（WA，TLE，MLE……）