E. Restorer Distance
Description
You have to restore the wall. The wall consists of NN pillars of bricks, the height of the ii-th pillar is initially equal to hihi, the height is measured in number of bricks. After the restoration all the NN pillars should have equal heights.
You are allowed the following operations:
- put a brick on top of one pillar, the cost of this operation is AA;
- remove a brick from the top of one non-empty pillar, the cost of this operation is RR;
- move a brick from the top of one non-empty pillar to the top of another pillar, the cost of this operation is MM.
You cannot create additional pillars or ignore some of pre-existing pillars even if their height becomes 00.
What is the minimal total cost of restoration, in other words, what is the minimal total cost to make all the pillars of equal height?
Input
The first line of input contains four integers NN, AA, RR, MM (1≤N≤1051≤N≤105, 0≤A,R,M≤1040≤A,R,M≤104) — the number of pillars and the costs of operations.
The second line contains NN integers hihi (0≤hi≤1090≤hi≤109) — initial heights of pillars.
Output
Print one integer — the minimal cost of restoration.
Examples
Input
3 1 100 1001 3 8
Output
12
正确解法：
有一堆砖，长度为hi。搬砖使他们的长度都相同。
hi++ 的代价为A，hi--的代价为R，把一个砖的一个搬到另一个砖的代价为M
M=min(M,A+R);
求最小代价。
仔细想想，假设最终砖的长度为x。
x增加的时候，代价会变多，x减少的时候，代价也会变多。
这就出现了三个峰，求最小代价。
三分。(学过但这是第一次用....)
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 #include <algorithm> 5 #include <set> 6 #include <queue> 7 #include <stack> 8 #include <string> 9 #include <cstring> 10 #include <vector> 11 #include <map> 12 //#include <unordered_map> 13 #define mem( a ,x ) memset( a , x ,sizeof(a) ) 14 #define rep( i ,x ,y ) for( int i = x ; i<=y ;i++ ) 15 #define lson l ,mid ,pos<<1 16 #define rson mid+1 ,r ,pos<<1|1 17 using namespace std; 18 typedef long long ll ; 19 typedef pair<int ,int> pii; 20 typedef pair<ll ,int> pli; 21 const int inf = 0x3f3f3f3f; 22 const ll mod=998244353; 23 const int N=1e5+50; 24 int T,n; 25 int A,R,M; 26 int s[N]; 27 ll check(int x) 28 { 29 ll suma=0,sumb=0; 30 for(int i=1;i<=n;i++) 31 { 32 if(s[i]<x) suma+=x-s[i]; 33 if(s[i]>x) sumb+=s[i]-x; 34 } 35 ll cnt=min(suma,sumb); 36 suma-=cnt; 37 sumb-=cnt; 38 return 1ll*suma*A+1ll*sumb*R+1ll*cnt*M; 39 } 40 int main() 41 { 42 scanf("%d%d%d%d",&n,&A,&R,&M); 43 M=min(M,A+R); 44 for(int i=1;i<=n;i++) 45 scanf("%d",s+i); 46 int l=0,r=1e9; 47 while(l<r) 48 { 49 int lmid=l+(r-l)/3,rmid=r-(r-l)/3; 50 if(check(lmid)<check(rmid)) r=rmid-1; 51 else l=lmid+1; 52 } 53 //cout<<l<<" "<<r<<endl; 54 printf("%lld ",check(l)); 55 56 return 0; 57 }
1 int l=0,r=1e9; 2 while(l<r) 3 { 4 int lmid=l+(r-l)/3,rmid=r-(r-l)/3; 5 if(check(lmid)<check(rmid)) r=rmid-1; 6 else l=lmid+1; 7 }