Skills CodeForces - 613B (双指针)

时间：2024-12-09 15:36:14

大意: $n$门课, 第$i$门分数$a_i$, 可以增加共$m$分, 求$cnt_{mx}*cf+mi*cm$的最大值 $cnt_{mx}$为满分的科目数, $mi$为最低分, $cf$, $cm$为给定系数

枚举满分的个数, 双指针求出最低分的最大值.

#include <iostream>

#include <sstream>

#include <algorithm>

#include <cstdio>

#include <math.h>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <string.h>

#include <bitset>

#define REP(i,a,n) for(int i=a;i<=n;++i)

#define PER(i,a,n) for(int i=n;i>=a;--i)

#define hr putchar(10)

#define pb push_back

#define lc (o<<1)

#define rc (lc|1)

#define mid ((l+r)>>1)

#define ls lc,l,mid

#define rs rc,mid+1,r

#define x first

#define y second

#define io std::ios::sync_with_stdio(false)

#define endl '\n'

#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})

using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

const int P = 1e9+7, INF = 0x3f3f3f3f;

ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}

ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}

inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}

//head

#ifdef ONLINE_JUDGE

const int N = 1e6+10;

#else

const int N = 111;

#endif

int n, A, cf, cm, Ans[N];

ll sum[N], m;

pii a[N];

struct {int mi,now,cnt;} opt;

int main() {

	scanf("%d%d%d%d%lld", &n, &A, &cf, &cm, &m);

	REP(i,1,n) scanf("%d", &a[i].x),a[i].y=i;

	sort(a+1,a+1+n);

	REP(i,1,n) sum[i]=sum[i-1]+a[i].x;

	ll ans = 0, res = m;

	int now = 1;

	while (now<n&&m+sum[now]>=(ll)now*a[now+1].x) ++now;

	int mi = min((ll)A,a[now].x+(m+sum[now]-(ll)now*a[now].x)/now);

	ans = (ll)mi*cm;

	opt={mi,now,0};

	REP(i,1,n) {

		res = m-((ll)i*A-sum[n]+sum[n-i]);

		if (res<0) break;

		now = min(now, n-i);

		while (now>1&&res+sum[now-1]<(ll)(now-1)*a[now].x) --now;

		mi = min((ll)A,now?a[now].x+(res+sum[now]-(ll)now*a[now].x)/now:INF);

		ll ret = (ll)i*cf+(ll)mi*cm;

		if (ret>ans) ans = ret, opt = {mi,now,i};

	}

	printf("%lld\n", ans);

	REP(i,1,opt.now) a[i].x=opt.mi;

	REP(i,n-opt.cnt+1,n) a[i].x=A;

	REP(i,1,n) Ans[a[i].y]=a[i].x;

	REP(i,1,n) printf("%d ", Ans[i]);hr;

}

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