#631187 (C++17) No.40 多項式の割り算

• 問題: No.40 多項式の割り算
• ユーザー: tanimani364
• 提出日時: 2021-03-18 17:32:25
• 言語: C++17 (gcc 13.3.0 + boost 1.87.0)
• 結果: WA
• 実行時間: -
• コード長: 2,142 bytes
• コンパイル時間: 1,872 ms
• コンパイル使用メモリ: 197,344 KB
• 最終ジャッジ日時: 2025-01-19 18:05:02
• ジャッジサーバーID （参考情報）: judge5 / judge5

ファイルパターン	結果
sample	AC * 3
other	AC * 21 WA * 11

ソースコード

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#include <bits/stdc++.h>
//#include<boost/multiprecision/cpp_int.hpp>
//#include<boost/multiprecision/cpp_dec_float.hpp>
//#include <atcoder/all>
#define rep(i, a) for (int i = (int)0; i < (int)a; ++i)
#define rrep(i, a) for (int i = (int)a - 1; i >= 0; --i)
#define REP(i, a, b) for (int i = (int)a; i < (int)b; ++i)
#define RREP(i, a, b) for (int i = (int)a - 1; i >= b; --i)
#define repl(i, a) for (ll i = (ll)0; i < (ll)a; ++i)
#define pb push_back
#define eb emplace_back
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define popcount __builtin_popcount
#define fi first
#define se second
using ll = long long;
constexpr ll mod = 1e9 + 7;
constexpr ll mod_998244353 = 998244353;
constexpr ll INF = 1LL << 60;
// #pragma GCC target("avx2")
// #pragma GCC optimize("O3")
// #pragma GCC optimize("unroll-loops")
//using lll=boost::multiprecision::cpp_int;
//using Double=boost::multiprecision::number<boost::multiprecision::cpp_dec_float<1024>>;//仮数部が1024桁
template <class T>
inline bool chmin(T &a, T b)
{
    if (a > b)
    {
        a = b;
        return true;
    }
    return false;
}
template <class T>
inline bool chmax(T &a, T b)
{
    if (a < b)
    {
        a = b;
        return true;
    }
    return false;
}
template <typename T>
T mypow(T x, T n, const T &p = -1)
{ //x^nをmodで割った余り
    if (p != -1)
    {
        x %= p;
    }
    T ret = 1;
    while (n > 0)
    {
        if (n & 1)
        {
            if (p != -1)
                ret = (ret * x) % p;
            else
                ret *= x;
        }
        if (p != -1)
            x = (x * x) % p;
        else
            x *= x;
        n >>= 1;
    }
    return ret;
}
using namespace std;
//using namespace atcoder;
void solve()
{
    int d;
    cin>>d;
    vector<int>a(d+1);
    rep(i,d+1)cin>>a[i];
    reverse(all(a));
    for(int i=0;i+2<d;++i){
        a[i+2]+=a[i];
        a[i]=0;
    }
    int pos=d+1;
    rep(i,d+1){
        if(a[i]>0){
            pos=i;
            break;
        }
    }
    vector<int>b;
    REP(i,pos,d+1){
        b.eb(a[i]);
    }
    if(b.size()==0){
        cout<<0<<"\n"<<0<<"\n";
        return;
    }
    reverse(all(b));
    int sz=b.size()-1;
    cout<<sz<<"\n";
    rep(i,b.size()){
        cout<<b[i]<<(i==b.size()-1?"\n":" ");
    }
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout << fixed << setprecision(15);
    solve();
    return 0;
}
 
 
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