题意：

思路：

与上一篇题解大同小异，无非就是不需要枚举排列了。

// Problem: E. String Reversal
// Contest: Codeforces - Educational Codeforces Round 96 (Rated for Div. 2)
// URL: https://codeforces.com/contest/1430/problem/E
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid (tr[u].l+tr[u].r>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;string s,ps;
vector<int>v[28];
struct Node {int l,r;LL cnt,lazy;
}tr[N<<2];void pushdown(int u) {if(!tr[u].lazy) return;LL lazy=tr[u].lazy; tr[u].lazy=0;tr[L].lazy+=lazy; tr[R].lazy+=lazy;tr[L].cnt+=lazy; tr[R].cnt+=lazy;
}void build(int u,int l,int r) {tr[u]={l,r,1,0};if(l==r) {tr[u].cnt=l;return;}build(L,l,Mid); build(R,Mid+1,r);
} void change(int u,int l,int r,int c) {if(tr[u].l>=l&&tr[u].r<=r) {tr[u].cnt+=c;tr[u].lazy+=c;return;}pushdown(u);if(l<=Mid) change(L,l,r,c);if(r>Mid) change(R,l,r,c);
}int query(int u,int l,int r) {if(tr[u].l>=l&&tr[u].r<=r) return tr[u].cnt;int ans=0;pushdown(u);if(l<=Mid) ans+=query(L,l,r);if(r>Mid) ans+=query(R,l,r);return ans;
}LL check() {LL ans=0;build(1,1,ps.length());for(int i=0;i<s.length();i++) {int pos=pos=v[s[i]-'a'].back(); v[s[i]-'a'].pop_back();LL now=query(1,pos,pos);ans+=abs(now-(i+1));change(1,1,pos,1);}return ans;
}int main()
{ios::sync_with_stdio(false);cin.tie(0);int _=1;while(_--) {int n; cin>>n;cin>>ps; s=ps; reverse(ps.begin(),ps.end());for(int i=ps.length()-1;i>=0;i--) {v[ps[i]-'a'].pb(i+1);}printf("%lld\n",check());}return 0;
}
/**/

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Educational Codeforces Round 96 E. String Reversa 线段树模拟序列交换

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题意：

思路：

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