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嘤嘤嘤，因为最近文化课老师追的紧了+班主任开班会，所以这博客是赶制的赝品

题目：

You‘ve got a string a1,a2,…,ana1,a2,…,an, consisting of zeros and ones.

Let‘s call a sequence of consecutive elements ai,ai + 1,…, ajai,ai + 1,…, aj (1≤ i≤ j≤ n1≤ i≤ j≤ n) a substring of string aa.

You can apply the following operations any number of times:

• Choose some substring of string aa (for example, you can choose entire string) and reverse it, paying xx coins for it (for example, «0101101» →→ «0111001»);
• Choose some substring of string aa (for example, you can choose entire string or just one symbol) and replace each symbol to the opposite one (zeros are replaced by ones, and ones — by zeros), paying yy coins for it (for example, «0101101» →→ «0110001»).

You can apply these operations in any order. It is allowed to apply the operations multiple times to the same substring.

What is the minimum number of coins you need to spend to get a string consisting only of ones?

Input

The first line of input contains integers nn, xx and yy (1 ≤ n ≤ 300000,0≤x,y≤1091 ≤ n ≤ 300000,0≤x,y≤109) — length of the string, cost of the first operation (substring reverse) and cost of the second operation (inverting all elements of substring).

The second line contains the string aa of length nn, consisting of zeros and ones.

Output

Print a single integer — the minimum total cost of operations you need to spend to get a string consisting only of ones. Print 00, if you do not need to perform any operations.

Examples

5 1 10
01000

11

#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
#include<map>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;
const int maxn=300000+5,INF=0x3f3f3f3f;
int n,m,c1,a[maxn],c2,low[maxn],f[maxn],len1,len0,len;
inline int read(){
   int s=0,w=1;
   char ch=getchar();
   while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)w=-1;ch=getchar();}
   while(ch>=‘0‘&&ch<=‘9‘) s=s*10+ch-‘0‘,ch=getchar();
   return s*w;
}
int main(){
    //freopen("a.in","r",stdin);
    n=read(),c1=read(),c2=read();
    for(int i=1;i<=n;i++){
        char x;
        cin>>x;
        x-=‘0‘;
        if(i==1||x!=a[len]){
            a[++len]=x;
            if(a[len]==1)len1++;
            else if(a[len]==0)len0++;
        }
    }
    if(len0==0){
        cout<<"0";
        return 0;
    }
    if(len1==0){
        cout<<c2;return 0;
    }
    int ans=0x3f;
    if(len1<len0)cout<<min((len0-1)*c1+c2,c2*len0);
    else if(len1>len0)cout<<min(len0*c1+c2,c2*len0);
    else cout<<min(len0*c1+c2,c2*len0);
}

 1 int main(){
 2     freopen("a.in","r",stdin);
 3     scanf("%d%d%d %s",&n,&c1,&c2,a+1);
 4     a[0]=‘1‘;//你问我为什么这里变成了scanf？因为提交多次发现卡死在了test4，我以为是快读beng了（遇到过卡快读的题）
 5     for(int i=1;i<=n;i++){
 6         if(a[i-1]==‘1‘&&a[i]==‘0‘)len0++;
 7     }
 8     if(len0==0){
 9         printf("0");
10     }else printf("%lld",(long long)min(c1*(len0-1)+c2,c2*len0);//卡在test4的原因：不开long long
11     return 0;
12 }

【盗版动归】Codeforces998C——Convert to Ones 归一操作

标签：row   etc   cond   stream   标题   int   ini   模拟   ORC   

原文地址：https://www.cnblogs.com/614685877--aakennes/p/12681643.html