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Colorful Slimes

简介: 题目描述 Snuke lives in another world, where slimes are real creatures and kept by some people. Slimes come in N colors. Those colors are conveniently numbered 1 through N. Snuke currently has no slime. His objective is to have slimes of all the colors together.
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题目描述


Snuke lives in another world, where slimes are real creatures and kept by some people. Slimes come in N colors. Those colors are conveniently numbered 1 through N. Snuke currently has no slime. His objective is to have slimes of all the colors together.


Snuke can perform the following two actions:


Select a color i (1≤i≤N), such that he does not currently have a slime in color i, and catch a slime in color i. This action takes him ai seconds.


Cast a spell, which changes the color of all the slimes that he currently has. The color of a slime in color i (1≤i≤N−1) will become color i+1, and the color of a slime in color N will become color 1. This action takes him x seconds.


Find the minimum time that Snuke needs to have slimes in all N colors.


Constraints

2≤N≤2,000

ai are integers.

1≤ai≤109

x is an integer.

1≤x≤109


输入


The input is given from Standard Input in the following format:


N x
a1 a2 … aN


输出


Find the minimum time that Snuke needs to have slimes in all N colors.


样例输入 Copy


2 10
1 100


样例输出 Copy


12


提示


Snuke can act as follows:


Catch a slime in color 1. This takes 1 second.

Cast the spell. The color of the slime changes: 1 → 2. This takes 10 seconds.

Catch a slime in color 1. This takes 1 second.

#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize (2)
#pragma G++ optimize (2)
#include <bits/stdc++.h>
#include <algorithm>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
#define wuyt main
typedef long long ll;
#define HEAP(...) priority_queue<__VA_ARGS__ >
#define heap(...) priority_queue<__VA_ARGS__,vector<__VA_ARGS__ >,greater<__VA_ARGS__ > >
template<class T> inline T min(T &x,const T &y){return x>y?y:x;}
template<class T> inline T max(T &x,const T &y){return x<y?y:x;}
ll read(){ll c = getchar(),Nig = 1,x = 0;while(!isdigit(c) && c!='-')c = getchar();
if(c == '-')Nig = -1,c = getchar();
while(isdigit(c))x = ((x<<1) + (x<<3)) + (c^'0'),c = getchar();
return Nig*x;}
#define read read()
const ll inf = 1e15;
const int maxn = 1e6 + 7;
const int mod = 1e9 + 7;
///ll n,m,k,x;
/**
ll superpow(ll a,ll b){
    ll ans=1;
    while (a>0)
    {
        if (a%2) ans=ans*b%n;
        b=b%n*b%n;
        a/=2;
    }
    return ans;
}**/
ll n,x;
ll num1[maxn],num2[maxn];
int main(){
    n=read,x=read;
    for(int i=0;i<n;i++) num1[i]=read;
    for(int i=0;i<n;i++) num2[i]=inf;
    ///memset(num2,0x3f3f3f3f,sizeof(num2));
    ll ans=inf;
    for(int k=0;k<n;k++){
        ll temp=k*x;
        for(int i=0;i<n;i++)
            num2[i]=min(num2[i],num1[(i-k+n)%n]);
        for(int j=0;j<n;j++)
            temp+=num2[j];
        ans=min(ans,temp);
    }
    cout<<ans<<endl;
    return 0;
}


ps:最小值设置为0x3f3f3f3f 本题不能通过


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