Codeforces 639D Bear and Contribution

Bear and Contribution

对于对于5余数为， 0, 1, 2, 3, 4的分别处理一次， 用优先队列贪心。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long

using namespace std;

const int N = 2e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

int n, k, b, c, t[N];
PII a[N];
LL ans = INF;

LL cnt = 0;
priority_queue<PLI> que;

LL solve(PII *a, LL b, LL c) {
    while(!que.empty()) que.pop();
    cnt = 0;
    LL ans = INF, ret = 0;
    for(int i = 1; i <= k; i++) {
        ret += a[i].se * c;
        ret += (a[i].fi - a[i - 1].fi) * cnt * b;
        que.push(mk(c * a[i].se + (a[1].fi - a[i].fi) * b, i));
        cnt++;
    }
    ans = min(ans, ret);
    for(int i = k + 1; i <= n; i++) {
        int p = que.top().se; que.pop(); cnt--;
        ret -= a[p].se * c;
        ret -= (a[i - 1].fi - a[p].fi) * b;
        ret += (a[i].fi - a[i - 1].fi) * cnt * b;
        ret += a[i].se * c;
        que.push(mk(c * a[i].se + (a[1].fi - a[i].fi) * b, i));
        cnt++;
        ans = min(ans, ret);
    }
    return ans;
}

int main() {
    scanf("%d%d%d%d", &n, &k, &b, &c);
    for(int i = 1; i <= n; i++) scanf("%d", &t[i]), t[i] += 1000000000;
    sort(t + 1, t + 1 + n);
    for(int i = 1; i <= n; i++) a[i].fi = t[i], a[i].se = 0;
    ans = min(ans, solve(a, c, 0));
    for(int j = 0; j < 5; j++) {
        for(int i = 1; i <= n; i++) {
            a[i].se = ((j - (t[i] % 5)) + 5) % 5;
            a[i].fi = (t[i] + a[i].se) / 5;
        }
        ans = min(ans, solve(a, b, c));
    }
    printf("%lld
", ans);
    return 0;
}

/*
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