D. Maximum Sum on Even Positions(翻转1次,求最大偶数位和)
作者:互联网
题:https://codeforces.com/contest/1373/problem/D
题意:翻转1次,求最大偶数位和
分析:先把偶数位的值加起来,然后只能翻转一次,那么要是有翻转奇数位的贡献只能是+a[i]-a[i-1]或+a[i]-a[i+1](这是对于翻转的整个子区间来说的),那么只要求贡献区间的最大子段和即可
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int mod=1e9+7; const int M=1e6+6; const int inf=0x3f3f3f3f; const ll INF=1e18; #define pb push_back #define MP make_pair ll a[M],maxx[M],b[M]; int main(){ int t; ios::sync_with_stdio(false); cin.tie(0); cin>>t; while(t--){ int n; cin>>n; ll ans=0; for(int i=1;i<=n;i++){ cin>>a[i]; if(i&1) ans+=a[i]; } int tot=0; for(int i=1;i<=n;i+=2) if(i+1<=n) b[++tot]=-a[i]+a[i+1]; maxx[1]=b[1]; ll tmp1=maxx[1]; for(int i=2;i<=tot;i++){ if(maxx[i-1]>0) maxx[i]=maxx[i-1]+b[i]; else maxx[i]=b[i]; tmp1=max(tmp1,maxx[i]); } /// tot=0; for(int i=1;i<=n;i+=2) if(i-1>=1) b[++tot]=-a[i]+a[i-1]; maxx[1]=b[1]; ll tmp2=maxx[1]; for(int i=2;i<=tot;i++){ if(maxx[i-1]>0) maxx[i]=maxx[i-1]+b[i]; else maxx[i]=b[i]; tmp2=max(tmp2,maxx[i]); } cout<<max(ans,max(tmp1+ans,tmp2+ans))<<endl; } return 0; }View Code
标签:Even,maxx,const,int,Positions,Sum,tmp2,ll,翻转 来源: https://www.cnblogs.com/starve/p/13284421.html