結果

問題 No.1825 Except One
ユーザー eQe
提出日時 2022-01-28 22:32:45
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 225 ms / 3,000 ms
コード長 3,836 bytes
コンパイル時間 4,095 ms
コンパイル使用メモリ 261,460 KB
最終ジャッジ日時 2025-01-27 16:59:22
ジャッジサーバーID
（参考情報） 		judge3 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 4
other AC * 31
権限があれば一括ダウンロードができます

ソースコード

diff # 

#include <atcoder/all>
using namespace atcoder;
#include <bits/stdc++.h>
#define all(a) (a.begin()),(a.end())
#define rall(a) (a.rbegin()),(a.rend())
using namespace std;
using ll=long long;
using ld=long double;
using pll=pair<ll,ll>;
template<class T>bool chmax(T &a, const T &b) {if(a<b) {a=b; return 1;} return 0;}
template<class T>bool chmin(T &a, const T &b) {if(b<a) {a=b; return 1;} return 0;}
istream& operator >> (istream& is, modint1000000007& x) { unsigned int t; is >> t; x=t; return is; }
istream& operator >> (istream& is, modint998244353& x) { unsigned int t; is >> t; x=t; return is; }
istream& operator >> (istream& is, modint& x) { unsigned int t; is >> t; x=t; return is; }
ostream& operator << (ostream& os, const modint1000000007& x) { os << x.val(); return os; }
ostream& operator << (ostream& os, const modint998244353& x) { os << x.val(); return os; }
ostream& operator << (ostream& os, const modint& x) { os << x.val(); return os; }
template<class T1, class T2>istream& operator >> (istream& is, pair<T1,T2>& p) { is >> p.first >> p.second; return is; }
template<class T1, class T2>ostream& operator << (ostream& os, const pair<T1,T2>& p) { os << p.first << " " << p.second; return os;}
template<class T>istream& operator >> (istream& is, vector<T>& v) { for(T& x:v) is >> x; return is; }
template<class T>ostream& operator << (ostream& os, const vector<T>& v) {for(int i=0;i<(int)v.size();i++) {os << v[i] << (i+1 == v.size() ? "":" ");} return os;}
template<class... A> void pt()      { std::cout << "\n"; }
template<class... A> void pt_rest() { std::cout << "\n"; }
template<class T, class... A> void pt_rest(const T& first, const A&... rest) { std::cout << " " << first; pt_rest(rest...); }
template<class T, class... A> void pt(const T& first, const A&... rest)      { std::cout << first; pt_rest(rest...); }
template<typename V, typename H> void resize(vector<V>& vec, const H head){ vec.resize(head); }
template<typename V, typename H, typename ... T> void resize(vector<V>& vec, const H& head, const T ... tail){ vec.resize(head); for(auto& v: vec) resize(v, tail...); }
template<typename V, typename T> void fill(V& x, const T& val){ x = val; }
template<typename V, typename T> void fill(vector<V>& vec, const T& val){ for(auto& v: vec) fill(v, val); }
template<typename H> void vin(istream& is, const int idx, vector<H>& head){ is >> head[idx]; }
template<typename H, typename ... T> void vin(istream& is, const int idx, vector<H>& head, T& ... tail){ vin(is >> head[idx], idx, tail...); }
template<typename H, typename ... T> void vin(istream& is, vector<H>& head, T& ... tail){ for(int i=0; i<(int)head.size(); i++) vin(is, i, head, tail...); }
template<typename H, typename ... T> void vin(vector<H>& head, T& ... tail){ vin(cin, head, tail...); }
map<ll,ll>divisor(ll n){map<ll,ll>res;for(ll i=1;i*i<=n;i++)if(n%i==0){res[i]++;res[n/i]++;}return res;}
map<ll,ll>factrization(ll n){map<ll,ll>res;for(ll i=2;i*i<=n;i++)while(n%i==0){res[i]++;n/=i;}if(n>1)res[n]++;return res;}
static const ll inf=1e18+7;
static const ld pi=acos(-1);
static const ld eps=1e-10;
using v1=vector<ll>;
using v2=vector<v1>;
using v3=vector<v2>;
using v4=vector<v3>;
using S=ll;
S op(S a,S b){return a+b;}
S e(){return 0;}


ll dp[50+1][5000+1][50+1][2];

int main(void) {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  
  ll N;cin>>N;
  v1 a(N);cin>>a;
  sort(all(a));
  ll S=accumulate(all(a),0LL);
  
  dp[0][0][0][1]=1;
  for(ll i=0;i<N;i++){
    for(ll j=0;j<=S;j++){
      for(ll k=0;k<=N;k++){
        dp[i+1][j][k][0]+=dp[i][j][k][0]+dp[i][j][k][1];
        if(j+a[i]<=S and k+1<=N)dp[i+1][j+a[i]][k+1][1]+=dp[i][j][k][0]+dp[i][j][k][1];
      }
    }
  }
  
  ll ans=0;
  for(ll i=1;i<=N;i++)for(ll j=0;j<=S;j++)for(ll k=2;k<=N;k++)
    if(j%(k-1)==0 and a[i-1]<=j/(k-1))ans+=dp[i][j][k][1];
  pt(ans);
}
0