2018-2019 Summer Petrozavodsk Camp, Oleksandr Kulkov Contest 2 Problem Solving Report

C

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Attendance problem, prefix and computing, all the negative items added to a set, if the current prefix and less than 0, from small to large pop-up set in negative terms, and the pop-up items added back to the prefix and.

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#include <bits/stdc++.h>

using namespace std;

#define ll long long
ll input(){
	ll x=0,f=0;char ch=getchar();
	while(ch<'0'||ch>'9') f|=ch=='-',ch=getchar();
	while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
	return f? -x:x;
}

#define PII pair<ll,int>
#define fr first
#define sc second
#define mp make_pair

const int N=2e5+7;

set<PII> s;
ll a[N],sum;
char Ans[N][20];

int main(){
	int n=input()+input();
	for(int i=1;i<=n;i++){
		a[i]=input();
		sum+=a[i];
		if(a[i]>0) strcpy(Ans[i],"resupplied");
		else{
			strcpy(Ans[i],"approved");
			s.insert(mp(-a[i],i));
		}
		while(sum<0){
			auto it=--s.end();
			sum+=it->fr;
			strcpy(Ans[it->sc],"declined");
			s.erase(it);
		}
	}
	for(int i=1;i<=n;i++){
		printf("%s\n",Ans[i]);
	}
}

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E

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And that the need to use pentagonal number theorem:

\[\prod_{n=1}^\infty(1-x^n)=\sum_{k=-\infty}^\infty(-1)^kx^{\frac{k(3k-1)}{2}}=\sum_{k=0}^\infty(-1)^kx^{\frac{k(3k\pm1)}{2}}\\ (1-x)(1-x^2)(1-x^3)\cdots=1-x-x^2+x^5+x^7-x^{12}-x^{15}+x^{22}+x^{26}\cdots \]

Coefficient known symbol relationships observed for the \ (+ - ++ - ++ - \) discussed before removing the three can be classified. For example, a paragraph is \ (+ X ^ A + X ^ BX ^ CX ^ D \) , then the \ (A \) position \ (. 1 \) , \ ([A +. 1, B] \) position \ (0 \) , \ ([B +. 1, C-. 1] \) position \ (. 9 \) , \ (C \) position \ (. 8 \) , \ ([C +. 1, D] \) location is \ (9 \) . Then by half a bit like a judge in that position.

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#include <bits/stdc++.h>

using namespace std;

#define ll long long
ll input(){
	ll x=0,f=0;char ch=getchar();
	while(ch<'0'||ch>'9') f|=ch=='-',ch=getchar();
	while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
	return f? -x:x;
}

ll f(ll k){return k*(k*3+1)/2;}

int main(){
	int T=input();
	while(T--){
		ll n=input();

		ll l=0,r=1e9,k;
		while(l<=r){
			ll mid=(l+r)>>1;
			if(f(mid)<n) k=mid,l=mid+1;
			else r=mid-1;
		}
		
		ll nr=f(k++);

		int Ans=0;
		if(k&1){
			if(n==nr+2*k-1) Ans=8;
			else if(n<=nr+3*k-1) Ans=9;
			else Ans=0;
		}else{
			if(n==nr+2*k-1) Ans=1;
			else if(n<=nr+3*k-1) Ans=0;
			else Ans=9;
		}
		printf("%d%c",Ans,T==0?'\n':' ');
	}
}