https://codeforces.com/contest/1144/problem/B

題意：給定一個(gè)數(shù)組，進(jìn)行操作：每次刪除一個(gè)數(shù)，但是刪除的數(shù)的奇偶性要與上一次相反。求刪除后剩下的數(shù)的最小和

題解：兩個(gè)隊(duì)列，先刪元素多的

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,p,l,r,u,v; int ans,cnt,flag,temp,sum; int a[N]; char str; struct node{}; int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){scanf("%d",&n);priority_queue<int>odd;priority_queue<int>even;for(int i=1;i<=n;i++){scanf("%d",&a[i]);if(a[i]%2){odd.push(a[i]);}else{even.push(a[i]);}sum+=a[i];}if(odd.empty()){cout<<sum-even.top()<<endl;return 0;}if(even.empty()){cout<<sum-odd.top()<<endl;return 0;}if(even.size()>odd.size()){flag=0;}else{flag=1;}while(1){if(flag&&!odd.empty()){sum-=odd.top();odd.pop();flag=!flag;}else if(!flag&&!even.empty()){sum-=even.top();even.pop();flag=!flag;}else{break;}}cout<<sum<<endl;//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

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