「杂题记录」出题人

📅 Last Updated: 2021-01-19
📊 Word Count: 363
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给出长度为 nn 的序列 aia_i ，需要构造一个序列 bib_i ，使得 ∀i,∃x,y∈[1,n],s.t.bx+by=ai\forall i, \exists x,y \in [1, n],s.t.b_x + b_y = a_i 要求输出序列 bib_i 和方案（aia_i 由哪两个 bjb_j 构成） 。无解输出 −1-1. d3A

n≤30n \le 30

分析

一道搜索模板题。

考虑如果能够构造出一个 aia_i 的一个子序列 a′a' 的对应长度的序列 b′b' 那么每次增加一个元素 aka_k 都可以直接构造 bkb_k。

如果 aia_i 存在偶数，可以直接构造 bi=ai2b_i = \frac{a_i}{2} 。其他的元素直接构造即可。

如果 aia_i 全部为奇数，考虑将 bib_i 中的元素看成点，aia_i 看成边，这样就有 nn 个点 nn 条边，一定存在一个环，只要找到这个环，剩下的元素就可以直接构造了

设：环上一个元素 bkb_k，那么对于任意一个环外的元素 ii ，可以直接构造成 bi=ai−bkb_i = a_i - b_k。

不妨设 环的大小为 kk， ai=bi+bi+1a_i = b_i + b_{i + 1}，ak=bk+b1a_k = b_k + b_1 。则：∑i=1kai=2∑i=1kbi\sum_{i=1}^{k}a_i = 2 \sum_{i = 1}^{k}b_i，易知： 2∣k2 | k。

可以直接拆成 a1+a3+⋯+ak−1=∑i=1kbi=a2+a4+⋯aka_1 + a_3 + \cdots+a_{k-1} = \sum_{i = 1}^{k}\limits{b_i}=a_2+a_4+\cdots a_k

传 统 艺 能 。直接搜索每个点 aia_i 是属于上式中的左边、右边 或者 不属于环上的点。要求满足左边的点权和等于右边的点权和。复杂度 O(3n)\mathcal{O}(3^n).

考虑把 aia_i 拆成两段，分别搜索，记录最终方案，然后线性合并。即：折半搜索。

复杂度 O(3n/2)\mathcal{O}(3^{n/2}) 。

赛后因为太菜不会线性合并，带了一个 log⁡\log 卡成了 30 体验很好。后来写 hash 挂链优化掉了一个 log⁡\log ，内存占用 2GiB⁡2\operatorname{GiB}，然后人就没了。至今任然 30。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <climits>
#include <map>
#include <vector>
#define GG() return (void)puts("No")
using namespace std;
#define LL long long
#define int long long 
const int _ = 1e2 + 5;
int n, A[_];

namespace subtask0{void work(){
    int target = 0;
    for(int i = 1; i <= n; i++) if(A[i] % 2) ; else { target = i; break; }
    cout << ("Yes") << endl;
    for(int i = 1; i <= n; i++){
        if(i == target) {
            cout << (A[i] / 2) << " ";
        } else {
            cout << (A[i] - (A[target] / 2)) << " ";
        }
    } cout << endl;
    for(int i = 1; i <= n; i++){
        if(i == target) {
            cout << target << " " << target << endl;
        } else {
            cout << target << " " << i << endl;
        }
    }
}}

namespace subtask1{


struct status_t{
    LL S;
    int A0; // the sum of.
    int A1; // the sum of.
    int sum0; // the number of.
    int sum1; // the number of.
    status_t() { sum0 = sum1 = S = A0 = A1 = 0; }
} ;

vector<status_t> reta, retb;

void d( int now, status_t st, const int & ed, vector<status_t> & ret){
    if(now > ed) {
        ret.push_back(st);
        return ;
    }
    status_t ext;
    ext = st; ext.S *= 3; ext.S += 0; ext.A0 += A[now]; ext.sum0 ++;
    d(now + 1, ext, ed, ret);
    
    ext = st; ext.S *= 3; ext.S += 1; ext.A1 += A[now]; ext.sum1 ++;
    d(now + 1, ext, ed, ret);
    
    ext = st; ext.S *= 3; ext.S += 2;
    d(now + 1, ext, ed, ret);
}



map<pair<int, int>, status_t> M;
#define none (-20031006)
int mid;

void output(status_t A, status_t B){
    vector<int> Node0, Node1, nodes;
    for(int i = 0, id = mid, tmp = A.S;       i < mid; i++, id--, tmp /= 3) { if(tmp % 3 == 0) Node0.push_back(id); if(tmp % 3 == 1) Node1.push_back(id); }
    for(int i = 0, id = n  , tmp = B.S; i < (n - mid); i++, id--, tmp /= 3) { if(tmp % 3 == 0) Node0.push_back(id); if(tmp % 3 == 1) Node1.push_back(id); }
    for(int i = 0; i < (int)Node0.size(); i++) nodes.push_back(Node0[i]), nodes.push_back(Node1[i]);
    static int Ans[_];  for(int i = 1; i <= n; i++) Ans[i] = none;
    static pair<int, int> Out[_];
    Ans[nodes[0]] = 0;
    for(int i = 0; i < (int)nodes.size(); i++){
        Ans[nodes[(i + 1) % nodes.size()]] = ::A[nodes[i]] - Ans[nodes[i]];
        Out[nodes[i]] = make_pair(nodes[i], nodes[(i + 1) % nodes.size()]);
    }
    static vector<int> ext;
    for(int i = 1; i <= n; i++){ if(Ans[i] != none) continue;
        Ans[i] = ::A[i];
        Out[i] = make_pair(nodes[0], i);
    }
    cout << ("Yes") << endl;
    for(int i = 1; i <= n; i++) cout << Ans[i] << " "; cout << endl;
    for(int i = 1; i <= n; i++) cout << Out[i].first << " " << Out[i].second << endl;
}


void work(){
    mid = (n >> 1);
    reta.clear(); retb.clear();
    d(1, status_t(), mid, reta);
    d(mid + 1, status_t(), n, retb);
    if(reta.size() == 0 || retb.size() == 0) GG();
    for(int i = 0; i < (int)reta.size(); i++) { pair<int, int> Key; Key = make_pair(reta[i].sum0 - reta[i].sum1, reta[i].A0 - reta[i].A1); M[Key] = reta[i]; }
    for(int i = 0; i < (int)retb.size(); i++) { 
        pair<int, int> Key; Key = make_pair(retb[i].sum1 - retb[i].sum0, retb[i].A1 - retb[i].A0);
        if(M.count(Key)) ; else continue;
        status_t A = M[Key], B = retb[i];
        if(A.sum0 + A.sum1 == 0 || B.sum0 + B.sum1 == 0) continue;
        if((A.sum0 + B.sum0 + A.sum1 + B.sum1) % 2) ;
        else return output(A, B);
    }
    GG();
} }
signed main(){
    // freopen(".in", "r", stdin);
    ios::sync_with_stdio(false);
    cin >> n; for(int i = 1; i <= n; i++) cin >> A[i];
    int pass = false;
    for(int i = 1; i <= n; i++) if(A[i] % 2) ; else { pass = true; break; }
    if(pass) subtask0::work();
    else subtask1::work();
    return 0;
}

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🔗 Source Hash: 4e283090a9bdb996f749953c850701eae302a4694376f1711a105bbb6c584677
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🔍 Filter Rules:
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    pdftitle={「杂题记录」出题人},
    pdfauthor={Jiayi Su (ShuYuMo)},
    hidelinks,
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  \title{「杂题记录」出题人}
  \author{Jiayi Su (ShuYuMo)}
  \date{2021-01-19 15:35:06}
  
  \begin{document}
  \maketitle
  
  \setstretch{1.3}
  给出长度为 \(n\) 的序列 \(a_i\) ，需要构造一个序列 \(b_i\) ，使得
  \(\forall i, \exists x,y \in [1, n],s.t.b_x + b_y = a_i\) 要求输出序列
  \(b_i\) 和方案（\(a_i\) 由哪两个 \(b_j\) 构成） 。无解输出 \(-1\). d3A
  
  \(n \le 30\)
  
  \subsection{分析}\label{ux5206ux6790}
  
  一道搜索\textbf{模板题}。
  
  考虑如果能够构造出一个 \(a_i\) 的一个子序列 \(a'\) 的对应长度的序列
  \(b'\) 那么每次增加一个元素 \(a_k\) 都可以直接构造 \(b_k\)。
  
  如果 \(a_i\) 存在偶数，可以直接构造 \(b_i = \frac{a_i}{2}\)
  。其他的元素直接构造即可。
  
  如果 \(a_i\) 全部为奇数，考虑将 \(b_i\) 中的元素看成点，\(a_i\)
  看成边，这样就有 \(n\) 个点 \(n\)
  条边，一定存在一个环，只要找到这个环，剩下的元素就可以直接构造了
  
  设：环上一个元素 \(b_k\)，那么对于任意一个环外的元素 \(i\)
  ，可以直接构造成 \(b_i = a_i - b_k\)。
  
  不妨设 环的大小为 \(k\)， \(a_i = b_i + b_{i + 1}\)，\(a_k = b_k + b_1\)
  。则：\(\sum_{i=1}^{k}a_i = 2 \sum_{i = 1}^{k}b_i\)，易知： \(2 | k\)。
  
  可以直接拆成
  \(a_1 + a_3 + \cdots+a_{k-1} = \sum_{i = 1}^{k}\limits{b_i}=a_2+a_4+\cdots a_k\)
  
  传 统 艺 能 。直接搜索每个点 \(a_i\) 是属于上式中的左边、右边 或者
  不属于环上的点。要求满足左边的点权和等于右边的点权和。复杂度
  \(\mathcal{O}(3^n)\).
  
  考虑把 \(a_i\)
  拆成两段，分别搜索，记录最终方案，然后线性合并。即：折半搜索。
  
  复杂度 \(\mathcal{O}(3^{n/2})\) 。
  
  赛后因为太菜不会线性合并，带了一个 \(\log\) 卡成了 30
  \st{体验很好}。后来写 hash 挂链优化掉了一个 \(\log\) ，内存占用
  \(2\operatorname{GiB}\)，然后人就没了。至今任然 30。
  
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  \NormalTok{    ext }\OperatorTok{=}\NormalTok{ st}\OperatorTok{;}\NormalTok{ ext}\OperatorTok{.}\NormalTok{S }\OperatorTok{*=} \DecValTok{3}\OperatorTok{;}\NormalTok{ ext}\OperatorTok{.}\NormalTok{S }\OperatorTok{+=} \DecValTok{1}\OperatorTok{;}\NormalTok{ ext}\OperatorTok{.}\NormalTok{A1 }\OperatorTok{+=}\NormalTok{ A}\OperatorTok{[}\NormalTok{now}\OperatorTok{];}\NormalTok{ ext}\OperatorTok{.}\NormalTok{sum1 }\OperatorTok{++;}
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  \NormalTok{    ext }\OperatorTok{=}\NormalTok{ st}\OperatorTok{;}\NormalTok{ ext}\OperatorTok{.}\NormalTok{S }\OperatorTok{*=} \DecValTok{3}\OperatorTok{;}\NormalTok{ ext}\OperatorTok{.}\NormalTok{S }\OperatorTok{+=} \DecValTok{2}\OperatorTok{;}
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  \PreprocessorTok{\#define none }\OperatorTok{({-}}\DecValTok{20031006}\OperatorTok{)}
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  \DataTypeTok{void}\NormalTok{ output}\OperatorTok{(}\DataTypeTok{status\_t}\NormalTok{ A}\OperatorTok{,} \DataTypeTok{status\_t}\NormalTok{ B}\OperatorTok{)\{}
  \NormalTok{    vector}\OperatorTok{\textless{}}\DataTypeTok{int}\OperatorTok{\textgreater{}}\NormalTok{ Node0}\OperatorTok{,}\NormalTok{ Node1}\OperatorTok{,}\NormalTok{ nodes}\OperatorTok{;}
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      \ControlFlowTok{for}\OperatorTok{(}\DataTypeTok{int}\NormalTok{ i }\OperatorTok{=} \DecValTok{0}\OperatorTok{,}\NormalTok{ id }\OperatorTok{=}\NormalTok{ n  }\OperatorTok{,}\NormalTok{ tmp }\OperatorTok{=}\NormalTok{ B}\OperatorTok{.}\NormalTok{S}\OperatorTok{;}\NormalTok{ i }\OperatorTok{\textless{}} \OperatorTok{(}\NormalTok{n }\OperatorTok{{-}}\NormalTok{ mid}\OperatorTok{);}\NormalTok{ i}\OperatorTok{++,}\NormalTok{ id}\OperatorTok{{-}{-},}\NormalTok{ tmp }\OperatorTok{/=} \DecValTok{3}\OperatorTok{)} \OperatorTok{\{} \ControlFlowTok{if}\OperatorTok{(}\NormalTok{tmp }\OperatorTok{\%} \DecValTok{3} \OperatorTok{==} \DecValTok{0}\OperatorTok{)}\NormalTok{ Node0}\OperatorTok{.}\NormalTok{push\_back}\OperatorTok{(}\NormalTok{id}\OperatorTok{);} \ControlFlowTok{if}\OperatorTok{(}\NormalTok{tmp }\OperatorTok{\%} \DecValTok{3} \OperatorTok{==} \DecValTok{1}\OperatorTok{)}\NormalTok{ Node1}\OperatorTok{.}\NormalTok{push\_back}\OperatorTok{(}\NormalTok{id}\OperatorTok{);} \OperatorTok{\}}
      \ControlFlowTok{for}\OperatorTok{(}\DataTypeTok{int}\NormalTok{ i }\OperatorTok{=} \DecValTok{0}\OperatorTok{;}\NormalTok{ i }\OperatorTok{\textless{}} \OperatorTok{(}\DataTypeTok{int}\OperatorTok{)}\NormalTok{Node0}\OperatorTok{.}\NormalTok{size}\OperatorTok{();}\NormalTok{ i}\OperatorTok{++)}\NormalTok{ nodes}\OperatorTok{.}\NormalTok{push\_back}\OperatorTok{(}\NormalTok{Node0}\OperatorTok{[}\NormalTok{i}\OperatorTok{]),}\NormalTok{ nodes}\OperatorTok{.}\NormalTok{push\_back}\OperatorTok{(}\NormalTok{Node1}\OperatorTok{[}\NormalTok{i}\OperatorTok{]);}
      \AttributeTok{static} \DataTypeTok{int}\NormalTok{ Ans}\OperatorTok{[}\NormalTok{\_}\OperatorTok{];}  \ControlFlowTok{for}\OperatorTok{(}\DataTypeTok{int}\NormalTok{ i }\OperatorTok{=} \DecValTok{1}\OperatorTok{;}\NormalTok{ i }\OperatorTok{\textless{}=}\NormalTok{ n}\OperatorTok{;}\NormalTok{ i}\OperatorTok{++)}\NormalTok{ Ans}\OperatorTok{[}\NormalTok{i}\OperatorTok{]} \OperatorTok{=}\NormalTok{ none}\OperatorTok{;}
      \AttributeTok{static}\NormalTok{ pair}\OperatorTok{\textless{}}\DataTypeTok{int}\OperatorTok{,} \DataTypeTok{int}\OperatorTok{\textgreater{}}\NormalTok{ Out}\OperatorTok{[}\NormalTok{\_}\OperatorTok{];}
  \NormalTok{    Ans}\OperatorTok{[}\NormalTok{nodes}\OperatorTok{[}\DecValTok{0}\OperatorTok{]]} \OperatorTok{=} \DecValTok{0}\OperatorTok{;}
      \ControlFlowTok{for}\OperatorTok{(}\DataTypeTok{int}\NormalTok{ i }\OperatorTok{=} \DecValTok{0}\OperatorTok{;}\NormalTok{ i }\OperatorTok{\textless{}} \OperatorTok{(}\DataTypeTok{int}\OperatorTok{)}\NormalTok{nodes}\OperatorTok{.}\NormalTok{size}\OperatorTok{();}\NormalTok{ i}\OperatorTok{++)\{}
  \NormalTok{        Ans}\OperatorTok{[}\NormalTok{nodes}\OperatorTok{[(}\NormalTok{i }\OperatorTok{+} \DecValTok{1}\OperatorTok{)} \OperatorTok{\%}\NormalTok{ nodes}\OperatorTok{.}\NormalTok{size}\OperatorTok{()]]} \OperatorTok{=} \OperatorTok{::}\NormalTok{A}\OperatorTok{[}\NormalTok{nodes}\OperatorTok{[}\NormalTok{i}\OperatorTok{]]} \OperatorTok{{-}}\NormalTok{ Ans}\OperatorTok{[}\NormalTok{nodes}\OperatorTok{[}\NormalTok{i}\OperatorTok{]];}
  \NormalTok{        Out}\OperatorTok{[}\NormalTok{nodes}\OperatorTok{[}\NormalTok{i}\OperatorTok{]]} \OperatorTok{=}\NormalTok{ make\_pair}\OperatorTok{(}\NormalTok{nodes}\OperatorTok{[}\NormalTok{i}\OperatorTok{],}\NormalTok{ nodes}\OperatorTok{[(}\NormalTok{i }\OperatorTok{+} \DecValTok{1}\OperatorTok{)} \OperatorTok{\%}\NormalTok{ nodes}\OperatorTok{.}\NormalTok{size}\OperatorTok{()]);}
      \OperatorTok{\}}
      \AttributeTok{static}\NormalTok{ vector}\OperatorTok{\textless{}}\DataTypeTok{int}\OperatorTok{\textgreater{}}\NormalTok{ ext}\OperatorTok{;}
      \ControlFlowTok{for}\OperatorTok{(}\DataTypeTok{int}\NormalTok{ i }\OperatorTok{=} \DecValTok{1}\OperatorTok{;}\NormalTok{ i }\OperatorTok{\textless{}=}\NormalTok{ n}\OperatorTok{;}\NormalTok{ i}\OperatorTok{++)\{} \ControlFlowTok{if}\OperatorTok{(}\NormalTok{Ans}\OperatorTok{[}\NormalTok{i}\OperatorTok{]} \OperatorTok{!=}\NormalTok{ none}\OperatorTok{)} \ControlFlowTok{continue}\OperatorTok{;}
  \NormalTok{        Ans}\OperatorTok{[}\NormalTok{i}\OperatorTok{]} \OperatorTok{=} \OperatorTok{::}\NormalTok{A}\OperatorTok{[}\NormalTok{i}\OperatorTok{];}
  \NormalTok{        Out}\OperatorTok{[}\NormalTok{i}\OperatorTok{]} \OperatorTok{=}\NormalTok{ make\_pair}\OperatorTok{(}\NormalTok{nodes}\OperatorTok{[}\DecValTok{0}\OperatorTok{],}\NormalTok{ i}\OperatorTok{);}
      \OperatorTok{\}}
  \NormalTok{    cout }\OperatorTok{\textless{}\textless{}} \OperatorTok{(}\StringTok{"Yes"}\OperatorTok{)} \OperatorTok{\textless{}\textless{}}\NormalTok{ endl}\OperatorTok{;}
      \ControlFlowTok{for}\OperatorTok{(}\DataTypeTok{int}\NormalTok{ i }\OperatorTok{=} \DecValTok{1}\OperatorTok{;}\NormalTok{ i }\OperatorTok{\textless{}=}\NormalTok{ n}\OperatorTok{;}\NormalTok{ i}\OperatorTok{++)}\NormalTok{ cout }\OperatorTok{\textless{}\textless{}}\NormalTok{ Ans}\OperatorTok{[}\NormalTok{i}\OperatorTok{]} \OperatorTok{\textless{}\textless{}} \StringTok{" "}\OperatorTok{;}\NormalTok{ cout }\OperatorTok{\textless{}\textless{}}\NormalTok{ endl}\OperatorTok{;}
      \ControlFlowTok{for}\OperatorTok{(}\DataTypeTok{int}\NormalTok{ i }\OperatorTok{=} \DecValTok{1}\OperatorTok{;}\NormalTok{ i }\OperatorTok{\textless{}=}\NormalTok{ n}\OperatorTok{;}\NormalTok{ i}\OperatorTok{++)}\NormalTok{ cout }\OperatorTok{\textless{}\textless{}}\NormalTok{ Out}\OperatorTok{[}\NormalTok{i}\OperatorTok{].}\NormalTok{first }\OperatorTok{\textless{}\textless{}} \StringTok{" "} \OperatorTok{\textless{}\textless{}}\NormalTok{ Out}\OperatorTok{[}\NormalTok{i}\OperatorTok{].}\NormalTok{second }\OperatorTok{\textless{}\textless{}}\NormalTok{ endl}\OperatorTok{;}
  \OperatorTok{\}}
  
  
  \DataTypeTok{void}\NormalTok{ work}\OperatorTok{()\{}
  \NormalTok{    mid }\OperatorTok{=} \OperatorTok{(}\NormalTok{n }\OperatorTok{\textgreater{}\textgreater{}} \DecValTok{1}\OperatorTok{);}
  \NormalTok{    reta}\OperatorTok{.}\NormalTok{clear}\OperatorTok{();}\NormalTok{ retb}\OperatorTok{.}\NormalTok{clear}\OperatorTok{();}
  \NormalTok{    d}\OperatorTok{(}\DecValTok{1}\OperatorTok{,} \DataTypeTok{status\_t}\OperatorTok{(),}\NormalTok{ mid}\OperatorTok{,}\NormalTok{ reta}\OperatorTok{);}
  \NormalTok{    d}\OperatorTok{(}\NormalTok{mid }\OperatorTok{+} \DecValTok{1}\OperatorTok{,} \DataTypeTok{status\_t}\OperatorTok{(),}\NormalTok{ n}\OperatorTok{,}\NormalTok{ retb}\OperatorTok{);}
      \ControlFlowTok{if}\OperatorTok{(}\NormalTok{reta}\OperatorTok{.}\NormalTok{size}\OperatorTok{()} \OperatorTok{==} \DecValTok{0} \OperatorTok{||}\NormalTok{ retb}\OperatorTok{.}\NormalTok{size}\OperatorTok{()} \OperatorTok{==} \DecValTok{0}\OperatorTok{)}\NormalTok{ GG}\OperatorTok{();}
      \ControlFlowTok{for}\OperatorTok{(}\DataTypeTok{int}\NormalTok{ i }\OperatorTok{=} \DecValTok{0}\OperatorTok{;}\NormalTok{ i }\OperatorTok{\textless{}} \OperatorTok{(}\DataTypeTok{int}\OperatorTok{)}\NormalTok{reta}\OperatorTok{.}\NormalTok{size}\OperatorTok{();}\NormalTok{ i}\OperatorTok{++)} \OperatorTok{\{}\NormalTok{ pair}\OperatorTok{\textless{}}\DataTypeTok{int}\OperatorTok{,} \DataTypeTok{int}\OperatorTok{\textgreater{}}\NormalTok{ Key}\OperatorTok{;}\NormalTok{ Key }\OperatorTok{=}\NormalTok{ make\_pair}\OperatorTok{(}\NormalTok{reta}\OperatorTok{[}\NormalTok{i}\OperatorTok{].}\NormalTok{sum0 }\OperatorTok{{-}}\NormalTok{ reta}\OperatorTok{[}\NormalTok{i}\OperatorTok{].}\NormalTok{sum1}\OperatorTok{,}\NormalTok{ reta}\OperatorTok{[}\NormalTok{i}\OperatorTok{].}\NormalTok{A0 }\OperatorTok{{-}}\NormalTok{ reta}\OperatorTok{[}\NormalTok{i}\OperatorTok{].}\NormalTok{A1}\OperatorTok{);}\NormalTok{ M}\OperatorTok{[}\NormalTok{Key}\OperatorTok{]} \OperatorTok{=}\NormalTok{ reta}\OperatorTok{[}\NormalTok{i}\OperatorTok{];} \OperatorTok{\}}
      \ControlFlowTok{for}\OperatorTok{(}\DataTypeTok{int}\NormalTok{ i }\OperatorTok{=} \DecValTok{0}\OperatorTok{;}\NormalTok{ i }\OperatorTok{\textless{}} \OperatorTok{(}\DataTypeTok{int}\OperatorTok{)}\NormalTok{retb}\OperatorTok{.}\NormalTok{size}\OperatorTok{();}\NormalTok{ i}\OperatorTok{++)} \OperatorTok{\{} 
  \NormalTok{        pair}\OperatorTok{\textless{}}\DataTypeTok{int}\OperatorTok{,} \DataTypeTok{int}\OperatorTok{\textgreater{}}\NormalTok{ Key}\OperatorTok{;}\NormalTok{ Key }\OperatorTok{=}\NormalTok{ make\_pair}\OperatorTok{(}\NormalTok{retb}\OperatorTok{[}\NormalTok{i}\OperatorTok{].}\NormalTok{sum1 }\OperatorTok{{-}}\NormalTok{ retb}\OperatorTok{[}\NormalTok{i}\OperatorTok{].}\NormalTok{sum0}\OperatorTok{,}\NormalTok{ retb}\OperatorTok{[}\NormalTok{i}\OperatorTok{].}\NormalTok{A1 }\OperatorTok{{-}}\NormalTok{ retb}\OperatorTok{[}\NormalTok{i}\OperatorTok{].}\NormalTok{A0}\OperatorTok{);}
          \ControlFlowTok{if}\OperatorTok{(}\NormalTok{M}\OperatorTok{.}\NormalTok{count}\OperatorTok{(}\NormalTok{Key}\OperatorTok{))} \OperatorTok{;} \ControlFlowTok{else} \ControlFlowTok{continue}\OperatorTok{;}
          \DataTypeTok{status\_t}\NormalTok{ A }\OperatorTok{=}\NormalTok{ M}\OperatorTok{[}\NormalTok{Key}\OperatorTok{],}\NormalTok{ B }\OperatorTok{=}\NormalTok{ retb}\OperatorTok{[}\NormalTok{i}\OperatorTok{];}
          \ControlFlowTok{if}\OperatorTok{(}\NormalTok{A}\OperatorTok{.}\NormalTok{sum0 }\OperatorTok{+}\NormalTok{ A}\OperatorTok{.}\NormalTok{sum1 }\OperatorTok{==} \DecValTok{0} \OperatorTok{||}\NormalTok{ B}\OperatorTok{.}\NormalTok{sum0 }\OperatorTok{+}\NormalTok{ B}\OperatorTok{.}\NormalTok{sum1 }\OperatorTok{==} \DecValTok{0}\OperatorTok{)} \ControlFlowTok{continue}\OperatorTok{;}
          \ControlFlowTok{if}\OperatorTok{((}\NormalTok{A}\OperatorTok{.}\NormalTok{sum0 }\OperatorTok{+}\NormalTok{ B}\OperatorTok{.}\NormalTok{sum0 }\OperatorTok{+}\NormalTok{ A}\OperatorTok{.}\NormalTok{sum1 }\OperatorTok{+}\NormalTok{ B}\OperatorTok{.}\NormalTok{sum1}\OperatorTok{)} \OperatorTok{\%} \DecValTok{2}\OperatorTok{)} \OperatorTok{;}
          \ControlFlowTok{else} \ControlFlowTok{return}\NormalTok{ output}\OperatorTok{(}\NormalTok{A}\OperatorTok{,}\NormalTok{ B}\OperatorTok{);}
      \OperatorTok{\}}
  \NormalTok{    GG}\OperatorTok{();}
  \OperatorTok{\}} \OperatorTok{\}}
  \DataTypeTok{signed}\NormalTok{ main}\OperatorTok{()\{}
      \CommentTok{// freopen(".in", "r", stdin);}
  \NormalTok{    ios}\OperatorTok{::}\NormalTok{sync\_with\_stdio}\OperatorTok{(}\KeywordTok{false}\OperatorTok{);}
  \NormalTok{    cin }\OperatorTok{\textgreater{}\textgreater{}}\NormalTok{ n}\OperatorTok{;} \ControlFlowTok{for}\OperatorTok{(}\DataTypeTok{int}\NormalTok{ i }\OperatorTok{=} \DecValTok{1}\OperatorTok{;}\NormalTok{ i }\OperatorTok{\textless{}=}\NormalTok{ n}\OperatorTok{;}\NormalTok{ i}\OperatorTok{++)}\NormalTok{ cin }\OperatorTok{\textgreater{}\textgreater{}}\NormalTok{ A}\OperatorTok{[}\NormalTok{i}\OperatorTok{];}
      \DataTypeTok{int}\NormalTok{ pass }\OperatorTok{=} \KeywordTok{false}\OperatorTok{;}
      \ControlFlowTok{for}\OperatorTok{(}\DataTypeTok{int}\NormalTok{ i }\OperatorTok{=} \DecValTok{1}\OperatorTok{;}\NormalTok{ i }\OperatorTok{\textless{}=}\NormalTok{ n}\OperatorTok{;}\NormalTok{ i}\OperatorTok{++)} \ControlFlowTok{if}\OperatorTok{(}\NormalTok{A}\OperatorTok{[}\NormalTok{i}\OperatorTok{]} \OperatorTok{\%} \DecValTok{2}\OperatorTok{)} \OperatorTok{;} \ControlFlowTok{else} \OperatorTok{\{}\NormalTok{ pass }\OperatorTok{=} \KeywordTok{true}\OperatorTok{;} \ControlFlowTok{break}\OperatorTok{;} \OperatorTok{\}}
      \ControlFlowTok{if}\OperatorTok{(}\NormalTok{pass}\OperatorTok{)}\NormalTok{ subtask0}\OperatorTok{::}\NormalTok{work}\OperatorTok{();}
      \ControlFlowTok{else}\NormalTok{ subtask1}\OperatorTok{::}\NormalTok{work}\OperatorTok{();}
      \ControlFlowTok{return} \DecValTok{0}\OperatorTok{;}
  \OperatorTok{\}}
  \end{Highlighting}
  \end{Shaded}
  
  
  \end{document}
[INFO] [makePDF] LaTeX run number 1
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[INFO] [makePDF] Rerun needed
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[INFO] [makePDF] LaTeX run number 2
[INFO] [makePDF] LaTeX output
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  (xeCJK)                
  (xeCJK)                Try to use `\setCJKmonofont[<...>]{<...>}' to define
  (xeCJK)                it.
  
  [1] [2] [3] [4] (.../input.aux)
  
  LaTeX Font Warning: Some font shapes were not available, defaults substituted.
  
   )
  Output written on .../input.pdf (4 pages).
  Transcript written on .../input.log.