【链接】 我是链接,点我呀:)

【题意】

在这里输入题意

【题解】

i从1..n/2循环一波。
保证a[i]和a[n-i+1]就好。
如果都是2的话填上min(a,b)*2就好
其他情况跟随非2的。

【代码】

#include <bits/stdc++.h>
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define all(x) x.begin(),x.end()
#define pb push_back
#define lson l,mid,rt<<1
#define ri(x) scanf("%d",&x)
#define rl(x) scanf("%lld",&x)
#define rs(x) scanf("%s",x)
#define rson mid+1,r,rt<<1|1
using namespace std; const double pi = acos(-1);
const int dx[4] = {0,0,1,-1};
const int dy[4] = {1,-1,0,0}; const int N = 20; int n,a,b,cost[2];
int c[N+10]; int main(){
#ifdef LOCAL_DEFINE
freopen("rush_in.txt", "r", stdin);
#endif
scanf("%d%d%d",&n,&a,&b);
cost[0] = a,cost[1] = b;
rep1(i,1,n) scanf("%d",&c[i]);
int ans = 0;
rep1(i,1,n/2){
int l = i,r = n-i+1;
if (c[l]==2 && c[r]==2){
ans+=min(a,b)*2;
}else{
if (c[l]==2){
ans+=cost[c[r]];
}else if (c[r]==2){
ans+=cost[c[l]];
}else if (c[l]!=c[r]){
cout<<-1<<endl;
return 0;
}
}
}
if (n&1){
if (c[n/2+1]==2){
ans+=min(a,b);
}
}
printf("%d\n",ans);
return 0;
}

【Codeforces Round #507 (Div. 2, based on Olympiad of Metropolises) A】Palindrome Dance的更多相关文章

  1. 【Codeforces Round #507 (Div. 2, based on Olympiad of Metropolises) B】Shashlik Cooking

    [链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 翻转一次最多影响2k+1个地方. 如果n<=k+1 那么放在1的位置就ok.因为能覆盖1..k+1 如果n<=2k+1 ...

  2. Codeforces Round #507 (Div. 2, based on Olympiad of Metropolises) D mt19937

    https://codeforces.com/contest/1040/problem/D 用法 mt19937 g(种子); //种子:time(0) mt19937_64 g(); //long ...

  3. Codeforces Round #433 (Div. 2, based on Olympiad of Metropolises)

    A. Fraction 题目链接:http://codeforces.com/contest/854/problem/A 题目意思:给出一个数n,求两个数a+b=n,且a/b不可约分,如果存在多组满足 ...

  4. Codeforces Round #433 (Div. 2, based on Olympiad of Metropolises) D. Jury Meeting(双指针模拟)

    D. Jury Meeting time limit per test 1 second memory limit per test 512 megabytes input standard inpu ...

  5. Codeforces Round #433 (Div. 2, based on Olympiad of Metropolises) D

    Country of Metropolia is holding Olympiad of Metrpolises soon. It mean that all jury members of the ...

  6. Codeforces Round #433 (Div. 2, based on Olympiad of Metropolises) C

    Helen works in Metropolis airport. She is responsible for creating a departure schedule. There are n ...

  7. Codeforces Round #433 (Div. 2, based on Olympiad of Metropolises) B

    Maxim wants to buy an apartment in a new house at Line Avenue of Metropolis. The house has n apartme ...

  8. Codeforces Round #433 (Div. 2, based on Olympiad of Metropolises) A

    Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned tha ...

  9. 【Codeforces Round #430 (Div. 2) A C D三个题】

    ·不论难度,A,C,D自己都有收获! [A. Kirill And The Game] ·全是英文题,述大意:    给出两组区间端点:l,r,x,y和一个k.(都是正整数,保证区间不为空),询问是否 ...

随机推荐

  1. UVA11234 Expressions

    题目的意思实在是读不懂,又是把栈变成队列什么的.. 只是大体的意思就是把后缀表达式变一下.. 抛开意思,事实上就是依据输入建个树,然后倒序输出.. 拿第一个例子说明:大写代表操作符(+ - × /之类 ...

  2. c18---数组和指针

    // // main.c // day09 #include <stdio.h> int main(int argc, const char * argv[]) { ; int *numP ...

  3. poj--2236--棋盘问题(dfs)

    棋盘问题 Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 31183   Accepted: 15469 Descriptio ...

  4. html5 初探

    html5是越来越火了.小小菜鸟也来学习学习. 相比于之前的几个版本,HTML5提供了更加丰富的多媒体标签使得音乐,视频的播放不用再借助于flah了.不过暂时各浏览器间样式还是有差别. 除此之外,表单 ...

  5. Spring配置之OpenSessionInViewFilter

    转自:https://www.cnblogs.com/blogonfly/articles/3991782.html 参考: OpenSessionInViewFilter作用及配置:http://w ...

  6. 6.11Realm简介

    CasRealm 统一认证授权中心  跟单点登录有关的.IniRealm 静态文件

  7. Wannafly挑战赛25 A 因子 数学

    题面 题意:令 X = n!,给定一大于1的正整数p,求一个k使得 p ^k | X 并且 p ^(k + 1) 不是X的因子,n,,p(1e18>=n>=1e4>=p>=2) ...

  8. Liunx搜索命令行

    1.grep grep(General Regular Expression Parser,通用规则表达式分析程序)是一种强大的文本搜索工具,它能使用正则表达式搜索文本,并把匹配的行打印出来. 它的使 ...

  9. [lua]异步串行流程*协程

    local function param_pack( params, callback ) table.insert(params, callback) return params end local ...

  10. spring IOC(DI)和AOP

    软件152谭智馗 IOC(Inversion of Control,控制倒转)Ioc意味着将你设计好的对象交给容器控制,而不是传统的在你的对象内部直接控制. DI—Dependency Injecti ...