牛客挑战赛34 2019-11-22

这个比赛太难了，本渣自然zero，喊一嗓子，还是博客园的界面我比较喜欢。牛客网的博客界面有点奇怪

第一题：

遇到了一个用时最短的神仙代码：

不仅模板类，内联函数，宏定义，编译器加速，using关键字用了很多C11新特性，代码狂拽酷炫吊炸天，我觉得看完他我就无法完成我的每日一题任务了，所以先贴在这，至少保证隔一天看一下，先看一天试试，下一次是11/28看这个代码

#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include <bits/stdc++.h>
#define fi first
#define se second
#define endl "\n"
using namespace std;
using db = double;
using ll = long long;
using ull = unsigned long long;
using pII = pair <int, int>;
using pLL = pair <ll, ll>;
constexpr int mod = 1e9 + 7;
template <class T1, class T2> inline void chadd(T1 &x, T2 y) { x += y; while (x >= mod) x -= mod; while (x < 0) x += mod; }
template <class T1, class T2> inline void chmax(T1 &x, T2 y) { if (x < y) x = y; }
template <class T1, class T2> inline void chmin(T1 &x, T2 y) { if (x > y) x = y; }
inline int rd() { int x; cin >> x; return x; }
template <class T> inline void rd(T &x) { cin >> x; }
template <class T> inline void rd(vector <T> &vec) { for (auto &it : vec) cin >> it; }
#define dbg(x...) do { cout << "\033[32;1m" << #x << " -> "; err(x); } while (0)
void err() { cout << "\033[39;0m" << endl; }
template <class T, class... Ts> void err(const T& arg, const Ts&... args) { cout << arg << ' '; err(args...); }
template <template<typename...> class T, typename t, typename... A>
void err(const T <t> &arg, const A&... args) { for (auto &v : arg) cout << v << ' '; err(args...); }
inline void pt() { cout << endl; }
template <class T, class... Ts> void pt(const T& arg, const Ts&... args) { cout << arg << ' '; pt(args...); }
template <template<typename...> class T, typename t, typename... A>
void pt(const T <t> &arg, const A&... args) { for (auto &v : arg) cout << v << ' '; pt(args...); }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
inline ll qpow(ll base, ll n) { ll res = 1; while (n) { if (n & 1) res = res * base % mod; base = base * base % mod; n >>= 1; } return res; }
//head
constexpr int N = 8e2 + 10;
int n, m, need, limit, a[N], p[N][N];
ll f[N][N];
void run() {
    for (int i = 1; i <= n; ++i) a[i] = rd();
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            p[i][j] = rd();
            if (j < limit) {
                p[i][j] += j * a[i];
            }
        }
    }
    memset(f, 0x3f, sizeof f);
    f[0][0] = 0;
    for (int i = 1; i <= n; ++i) {
        for (int j = 0; j <= need; ++j) f[i][j] = f[i - 1][j];
        for (int j = 1; j <= min(m, need); ++j) {
            for (int k = need; k >= j; --k) {
                chmin(f[i][k], f[i - 1][k - j] + p[i][j]);
            }
        }
    }
    pt(f[n][need]);
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);
    cout << fixed << setprecision(20);
    while (cin >> n >> m >> need >> limit) run();
    return 0;
}