Codeforces-855B - Marvolo Gaunt's Ring - DP
2018-01-23 17:25
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题目:
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
input5 1 2 3 1 2 3 4 5
output
30
input
5 1 2 -3 -1 -2 -3 -4 -5
output
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.
题意:
给出p,q,r,给出n个数v[i],问max(p∗ai+q∗aj+r∗ak) (i≤j≤k)思路:
枚举aj,当0≤p,选取max(ai),i∈[1,j], 否则选取min(ai),i∈[1,j]。q同理。对于取max和取min,维护前缀和后缀即可。
实现:
#include <bits/stdc++.h> using namespace std; #define ll long long const ll maxn = int(1e5)+7, INF = 0x7f7f7f7f7f7f7f7f; ll n, p, q, r; ll premi[maxn], prema[maxn], sufmi[maxn], sufma[maxn], num[maxn]; int main() { scanf("%lld%lld%lld%lld", &n, &p, &q, &r); for(int i=1 ; i<=n ; i++) scanf("%lld", num+i); prema[0] = sufma[n+1] = -INF; premi[0] = sufmi[n+1] = INF; for(int i=n ; i>=1 ; i--) { sufma[i] = max(num[i], sufma[i+1]); sufmi[i] = min(num[i], sufmi[i+1]); } for(int i=1 ; i<=n ; i++) { prema[i] = max(num[i], prema[i-1]); premi[i] = min(num[i], premi[i-1]); } ll ans = -INF, tmp; for(int i=1 ; tmp = 0, i<=n ; i++) { tmp += p * (p < 0 ? premi[i] : prema[i]); tmp += r * (r < 0 ? sufmi[i] : sufma[i]); ans = max(ans, tmp + q * num[i]); } printf("%lld\n", ans); return 0; }
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