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 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + ;
const double eps = 1e-;
const double PI = acos(-); int n, k, b, c, t[N];
PII a[N];
LL ans = INF; LL cnt = ;
priority_queue<PLI> que; LL solve(PII *a, LL b, LL c) {
while(!que.empty()) que.pop();
cnt = ;
LL ans = INF, ret = ;
for(int i = ; i <= k; i++) {
ret += a[i].se * c;
ret += (a[i].fi - a[i - ].fi) * cnt * b;
que.push(mk(c * a[i].se + (a[].fi - a[i].fi) * b, i));
cnt++;
}
ans = min(ans, ret);
for(int i = k + ; i <= n; i++) {
int p = que.top().se; que.pop(); cnt--;
ret -= a[p].se * c;
ret -= (a[i - ].fi - a[p].fi) * b;
ret += (a[i].fi - a[i - ].fi) * cnt * b;
ret += a[i].se * c;
que.push(mk(c * a[i].se + (a[].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 = ; i <= n; i++) scanf("%d", &t[i]), t[i] += ;
sort(t + , t + + n);
for(int i = ; i <= n; i++) a[i].fi = t[i], a[i].se = ;
ans = min(ans, solve(a, c, ));
for(int j = ; j < ; j++) {
for(int i = ; i <= n; i++) {
a[i].se = ((j - (t[i] % )) + ) % ;
a[i].fi = (t[i] + a[i].se) / ;
}
ans = min(ans, solve(a, b, c));
}
printf("%lld\n", ans);
return ;
} /*
*/
上一篇:编码(2)从字节理解Unicode(UTF8/UTF16)


下一篇:Go语言之高级篇beego框架之日志收集系统