結果

問題 No.664 超能力者Aと株価予測
ユーザー mai
提出日時 2018-03-09 23:52:23
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 19 ms / 4,000 ms
コード長 4,640 bytes
コンパイル時間 3,241 ms
コンパイル使用メモリ 215,820 KB
最終ジャッジ日時 2025-01-05 09:12:11
ジャッジサーバーID
(参考情報)
judge3 / judge4
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ファイルパターン 結果
other AC * 15
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 #
プレゼンテーションモードにする

#pragma GCC optimize ("O3")
#pragma GCC target ("avx")
#include "bits/stdc++.h" // define macro "/D__MAI"
using namespace std;
typedef long long int ll;
#define debug(v) {printf("L%d %s > ",__LINE__,#v);cout<<(v)<<endl;}
#define debugv(v) {printf("L%d %s > ",__LINE__,#v);for(auto e:(v)){cout<<e<<" ";}cout<<endl;}
#define debuga(m,w) {printf("L%d %s > ",__LINE__,#m);for(int x=0;x<(w);x++){cout<<(m)[x]<<" ";}cout<<endl;}
#define debugaa(m,h,w) {printf("L%d %s >\n",__LINE__,#m);for(int y=0;y<(h);y++){for(int x=0;x<(w);x++){cout<<(m)[y][x]<<" ";}cout<<endl;}}
#define ALL(v) (v).begin(),(v).end()
#define repeat(cnt,l) for(auto cnt=0ll;(cnt)<(l);++(cnt))
#define rrepeat(cnt,l) for(auto cnt=(l)-1;0<=(cnt);--(cnt))
#define iterate(cnt,b,e) for(auto cnt=(b);(cnt)!=(e);++(cnt))
#define diterate(cnt,b,e) for(auto cnt=(b);(cnt)!=(e);--(cnt))
#define MD 1000000007ll
#define PI 3.1415926535897932384626433832795
template<typename T1, typename T2> ostream& operator <<(ostream &o, const pair<T1, T2> p) { o << "(" << p.first << ":" << p.second << ")"; return o;
    }
template<typename T> T& maxset(T& to, const T& val) { return to = max(to, val); }
template<typename T> T& minset(T& to, const T& val) { return to = min(to, val); }
void bye(string s, int code = 0) { cout << s << endl; exit(code); }
mt19937_64 randdev(8901016);
inline ll rand_range(ll l, ll h) {
return uniform_int_distribution<ll>(l, h)(randdev);
}
#if defined(_WIN32) || defined(_WIN64)
#define getchar_unlocked _getchar_nolock
#define putchar_unlocked _putchar_nolock
#elif defined(__GNUC__)
#else
#define getchar_unlocked getchar
#define putchar_unlocked putchar
#endif
namespace {
#define isvisiblechar(c) (0x21<=(c)&&(c)<=0x7E)
class MaiScanner {
public:
template<typename T> void input_integer(T& var) {
var = 0; T sign = 1;
int cc = getchar_unlocked();
for (; cc<'0' || '9'<cc; cc = getchar_unlocked())
if (cc == '-') sign = -1;
for (; '0' <= cc && cc <= '9'; cc = getchar_unlocked())
var = (var << 3) + (var << 1) + cc - '0';
var = var * sign;
}
inline int c() { return getchar_unlocked(); }
inline MaiScanner& operator>>(int& var) { input_integer<int>(var); return *this; }
inline MaiScanner& operator>>(long long& var) { input_integer<long long>(var); return *this; }
inline MaiScanner& operator>>(string& var) {
int cc = getchar_unlocked();
for (; !isvisiblechar(cc); cc = getchar_unlocked());
for (; isvisiblechar(cc); cc = getchar_unlocked())
var.push_back(cc);
return *this;
}
template<typename IT> void in(IT begin, IT end) { for (auto it = begin; it != end; ++it) *this >> *it; }
};
class MaiPrinter {
public:
template<typename T>
void output_integer(T var) {
if (var == 0) { putchar_unlocked('0'); return; }
if (var < 0)
putchar_unlocked('-'),
var = -var;
char stack[32]; int stack_p = 0;
while (var)
stack[stack_p++] = '0' + (var % 10),
var /= 10;
while (stack_p)
putchar_unlocked(stack[--stack_p]);
}
inline MaiPrinter& operator<<(char c) { putchar_unlocked(c); return *this; }
inline MaiPrinter& operator<<(int var) { output_integer<int>(var); return *this; }
inline MaiPrinter& operator<<(long long var) { output_integer<long long>(var); return *this; }
inline MaiPrinter& operator<<(char* str_p) { while (*str_p) putchar_unlocked(*(str_p++)); return *this; }
inline MaiPrinter& operator<<(const string& str) {
const char* p = str.c_str();
const char* l = p + str.size();
while (p < l) putchar_unlocked(*p++);
return *this;
}
template<typename IT> void join(IT begin, IT end, char sep = '\n') { for (auto it = begin; it != end; ++it) *this << *it << sep; }
};
}
MaiScanner scanner;
MaiPrinter printer;
ll m, n, kei;
ll dp[400][22];
ll aa[400];
int main() {
scanner >> n >> m >> kei;
++n;
scanner.in(aa, aa + n);
ll best = kei;
dp[0][0] = kei;
iterate(i, 1, n) {
repeat(j, i) {
rrepeat(k, m) {
maxset(dp[i][k], dp[j][k]);
maxset(best,
maxset(dp[i][k + 1], dp[j][k] / aa[j] * aa[i] + (dp[j][k] % aa[j]))
);
}
}
}
//debugaa(dp, 10,10);
cout << best << endl;
return 0;
}
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