标签：amp i++ flag queue const title imu des string
Professor Dumbledore is helping Harry destroy the Horcruxes. He went to Gaunt Shack as he suspected a Horcrux to be present there. He saw Marvolo Gaunt‘s Ring and identified it as a Horcrux. Although he destroyed it, he is still affected by its curse. Professor Snape is helping Dumbledore remove the curse. For this, he wants to give Dumbledore exactly x drops of the potion he made.
Value of x is calculated as maximum of p·ai + q·aj + r·ak for given p, q, r and array a1, a2, ... an such that 1 ≤ i ≤ j ≤ k ≤ n. Help Snape find the value of x. Do note that the value of x may be negative.
Input
First line of input contains 4 integers n, p, q, r ( - 109 ≤ p, q, r ≤ 109, 1 ≤ n ≤ 105).
Next line of input contains n space separated integers a1, a2, ... an ( - 109 ≤ ai ≤ 109).
Output
Output a single integer the maximum value of p·ai + q·aj + r·ak that can be obtained provided 1 ≤ i ≤ j ≤ k ≤ n.
Examples
5 1 2 3
1 2 3 4 5
30
5 1 2 -3
-1 -2 -3 -4 -5
12
Note
In the first sample case, we can take i = j = k = 5, thus making the answer as 1·5 + 2·5 + 3·5 = 30.
In second sample case, selecting i = j = 1 and k = 5 gives the answer 12.
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <string>
#include <cstring>
#include <cstdlib>
#include <map>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
typedef long long lli;
using namespace std;
const int mxn = 1e9;
#define TLE std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); cout.precision(10);
multiset<int> v;
multiset<int> :: iterator it ;
int main()
{
TLE;
lli n,cnt=1,m,k,ll;
while(cin>>n>>m>>k>>ll)
{
int flag1 = 0,flag2=0;lli mx,x;
if(m>=0 && k>=0 && ll>=0 )
flag1=1;
if(m<0 && k<0 && ll<0 )
flag2=1;
if(flag1)
{
cin>>mx;
for(int i=2; i<=n; i++)
{
cin>>x;
mx = max( mx ,x );
}
cout<<(m+k+ll)*mx<<endl;
}
else if(flag2)
{
cin>>mx;
for(int i=2; i<=n; i++)
{
cin>>x;
mx = min( mx ,x );
}
cout<<(m+k+ll)*mx<<endl;
}
else
{
lli dp[n+5];
lli l[n+5],r[n+5];
for(int i=1;i<=n;i++)
{
cin>>dp[i];
}
l[1] = m*dp[1];
for(int i=2; i<=n; i++)
{
l[i] = max( l[i-1] ,m*dp[i] );
}
r[n] = ll*dp[n];
for(int i=n-1; i>=1; i--)
{
r[i] = max( r[i+1] , ll*dp[i] );
}
mx = -1000000000000000000000-5;
for(int i=1; i<=n; i++)
mx = max( l[i]+r[i]+dp[i]*k , mx);
cout<<mx<<endl;
}
}
return 0;
}
标签：amp i++ flag queue const title imu des string
原文地址：https://www.cnblogs.com/Shallow-dream/p/11623521.html
