题解：AT_abc393_d [ABC393D] Swap to Gather

/*
#include<bits/stdc++.h>
using namespace std;
#define int long long
int n,a[500010],sum[500010],Sum,cnt,cnt1,ans=10000000000000000;

signed main(){
	cin>>n;
	for(int i=1;i<=n;i++){
		char tmp;
		cin>>tmp;
		a[i]=(int)(tmp-'0');
		sum[i]=sum[i-1]+(int)(a[i]==0),cnt1+=(int)(a[i]==1);
	}
	for(int i=1;i<=n;i++){
		if(a[i])
			Sum+=sum[i];
	}
	int qwq=0;
	for(int i=1;i<=n;i++){
		int tmp=0;
		if(a[i]==1)
			qwq+=sum[i],cnt++;
		tmp=cnt*sum[i]-qwq+Sum-qwq-(cnt1-cnt)*sum[i];
		ans=min(ans,tmp);
	}
	cout<<ans;
}