結果

問題 No.1884 Sequence
ユーザー woodywoody
提出日時 2022-06-15 08:14:47
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 136 ms / 2,000 ms
コード長 3,459 bytes
コンパイル時間 3,898 ms
コンパイル使用メモリ 233,956 KB
実行使用メモリ 6,820 KB
最終ジャッジ日時 2024-10-03 16:10:38
合計ジャッジ時間 8,479 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge1
このコードへのチャレンジ
（要ログイン）
ファイルパターン 結果
sample AC * 3
other AC * 40
権限があれば一括ダウンロードができます
コンパイルメッセージ
In function 'll gcd(ll, ll)',
    inlined from 'int main()' at main.cpp:113:26:
main.cpp:61:30: warning: 'diffgcd' may be used uninitialized [-Wmaybe-uninitialized]
   61 | ll gcd(ll a, ll b) { return a?gcd(b%a,a):b;}
      |                             ~^~~~~~~~~~~~~
main.cpp: In function 'int main()':
main.cpp:104:8: note: 'diffgcd' was declared here
  104 |     ll diffgcd;
      |        ^~~~~~~

ソースコード

diff # 

#include<bits/stdc++.h>
#include<atcoder/all>
#define rep(i,b) for(int i=0;i<b;i++)
#define rrep(i,b) for(int i=b-1;i>=0;i--)
#define rep1(i,b) for(int i=1;i<b;i++)
#define repx(i,x,b) for(int i=x;i<b;i++)
#define fore(i,a) for(auto i:a)
#define fore1(i,a) for(auto &i:a)
#define rng(x) (x).begin(), (x).end()
#define rrng(x) (x).rbegin(), (x).rend()
#define sz(x) ((int)(x).size())
#define pb push_back
#define fi first
#define se second

using namespace std;
using namespace atcoder;

using ll = long long;
template<typename T> using mpq = priority_queue<T, vector<T>, greater<T>>;
template<typename T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<typename T> bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
template<typename T> ll sumv(const vector<T>&a){ll res(0);for(auto&&x:a)res+=x;return res;}
bool yn(bool a) { if(a) {cout << "Yes" << endl; return 1;} else {cout << "No" << endl; return 0;}}
#define dame { cout << "No" << endl; return 0;}
#define dame1 { cout << -1 << endl; return 0;}
#define test(x) cout << "test" << x << endl;
#define deb(x,y) cout << x << " " << y << endl;
#define deb3(x,y,z) cout << x << " " << y << " " << z << endl;
#define deb4(x,y,z,x2) cout << x << " " << y << " " << z << " " << x2 << endl;
#define out cout << ans << endl;
#define outv fore(yans , ans) cout << yans << "\n";
#define show(x) cerr<<#x<<" = "<<x<<endl;
#define showv(v) {fore(vy , v) {cout << vy << " ";} cout << endl;}
#define showv2(v) fore(vy , v) cout << vy << "\n";
using pll = pair<ll,ll>;
using pil = pair<int,ll>;
using pli = pair<ll,int>;
using pii = pair<int,int>;
using tp = tuple<int ,int ,int>;
using vi = vector<int>;
using vl = vector<ll>;
using vs = vector<string>;
using vb = vector<bool>;
using vpii = vector<pii>;
using vpli = vector<pli>;
using vpll = vector<pll>;
using vpil = vector<pil>;
using vvi = vector<vector<int>>;
using vvl = vector<vector<ll>>;
using vvs = vector<vector<string>>;
using vvb = vector<vector<bool>>;
using vvpii = vector<vector<pii>>;
using vvpli = vector<vector<pli>>;
using vvpll = vector<vpll>;
using vvpil = vector<vpil>;
using mint = modint998244353;
using vm = vector<mint>;
using vvm = vector<vector<mint>>;
vector<int> dx={1,0,-1,0,1,1,-1,-1},dy={0,1,0,-1,1,-1,1,-1};
ll gcd(ll a, ll b) { return a?gcd(b%a,a):b;}
ll lcm(ll a, ll b) { return a/gcd(a,b)*b;}
const double eps = 1e-10;
const ll LINF = 1001002003004005006ll;
const int INF = 1001001001;

int main(){
    int n; cin>>n;
    vl a(n);

    bool ok = true;
    rep(i,n){
        cin>>a[i];
        if (a[i] == 0) ok = false;
    }

    if (ok){
        sort(rng(a));
        ll diff = a[1] - a[0];

        ok = true;

        rep(i,n-1) if (a[i+1] - a[i] != diff) ok = false;

        yn(ok);

        return 0;
    }

    ll zero = 0;
   
    sort(rng(a));

    while(zero <= n-1 && a[zero] == 0) zero++;
    
    if (zero >= n-2){
        cout << "Yes" << endl;

        return 0;
    }

    ll l = zero;

    ll diffgcd;

    ok = true;

    repx(i,l,n-1){
        ll tmp = a[i+1] - a[i];
        if (tmp == 0) ok = false;
        if (i == l) diffgcd = tmp;
        else{
            diffgcd = gcd(diffgcd , tmp);
        }
    }

    if (!ok){
        ok = true;
        repx(i,l,n-1) if (a[i+1] != a[i]) ok = false;

        yn(ok);

        return 0;
    }

    ll cnt = 0;

    repx(i,l,n-1) cnt += ((a[i+1] - a[i]) / diffgcd) - 1;

    yn(cnt <= zero);

    

    return 0;
}
0