（思维+STL）Gym - 101669F

题意：将 串1 ：  01110   变成串2 ：   10011；

           变化规则：将0变成1或将1变成0 ，之后值为串1中1的vel和

           0 1 1 1 0

           0 1 0 1 0      3     和为 3

           0 0 0 1 0      1     和为3+1

           1 0 0 1 0      6     和为3+1+6

           1 0 0 1 1      11    和为3+1+6+11=21

#include<set>
#include<map>
#include<list>
#include<queue>
#include<stack>
#include<math.h>
#include<vector>
#include<bitset>
#include<iomanip>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#define eps (1e-8)
#define MAX 0x3f3f3f3f
#define u_max 1844674407370955161
#define l_max 9223372036854775807
#define i_max 2147483647
#define re register
#define pushup() tree[rt]=tree[rt<<1]+tree[rt<<1|1]
#define nth(k,n) nth_element(a,a+k,a+n);  // 将 第K大的放在k位
#define ko() for(int i=2;i<=n;i++) s=(s+k)%i // 约瑟夫
#define ok() v.erase(unique(v.begin(),v.end()),v.end()) // 排序,离散化
#define Catalan C(2n,n)-C(2n,n-1)  (1,2,5,14,42,132,429...) // 卡特兰数
using namespace std;

inline int read(){
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' & c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}

typedef long long ll;
const double pi = atan(1.)*4.;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3fLL;
const int M=63;
const int N=1e5+5;
ll a[N],b[N];
char s1[N],s2[N];
multiset<ll>ss1,ss2;

bool cmp(ll x,ll y){
    return x>y;
}
int main(){
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%lld",&a[i]);
    }
    scanf(" %s",s1);
    scanf(" %s",s2);
    int p=0; ll sum=0,sum1=0;
    for(int i=0;i<n;i++){
        if(s1[i]==s2[i]&&s1[i]=='1'){
            b[p++]=a[i];
            sum1+=a[i];
            sum+=a[i];
        }
        if(s1[i]!=s2[i]){
            if(s1[i]=='1'){
                sum+=a[i];
                ss1.insert(a[i]);
            }
            else
                ss2.insert(a[i]);
        }
    }
    sort(b,b+p,cmp);
    multiset<ll>::reverse_iterator it1;
    multiset<ll>::iterator it2;
    ll ans=INF;
    for(int i=-1;i<p;i++){
        if(i!=-1){
            ss1.insert(b[i]);
            sum1-=b[i];
            ss2.insert(b[i]);
        }
        ll cut=sum,k=0;
        for(it1=ss1.rbegin();it1!=ss1.rend();it1++){
            cut-=(*it1);
            //printf("cut == %lld\n",cut);
            k+=cut;
        }
        cut=sum1;
        for(it2=ss2.begin();it2!=ss2.end();it2++){
            cut+=(*it2);
            //printf("cut == %lld\n",cut);
            k+=cut;
        }
        //printf("k  ==  %lld\n",k);
        ans=min(ans,k);
    }
    printf("%lld\n",ans);
    return 0;
}