以下代码：

cl数组：到达i时共有多少个朝向左边的箭头

cr数组：到达i时共有多少个朝向右边的箭头

sum_l：到达i时所有朝向左侧的箭头的下标和

sum_r：到达i时所有朝向右侧的箭头的下标和

以下代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <string.h>
using namespace std ;
int t , n , pl , pr ,pre_l[500005] , pre_r[500005] , sum_l[500005] , sum_r[500005] ;
int main(){
    cin >> t ;
    while( t -- ){
        pl = pr = 1 ;
        memset(pre_l , 0 , sizeof(pre_l));
        memset(pre_r , 0 , sizeof(pre_r));
        memset(sum_l , 0 , sizeof(sum_l));
        memset(sum_r , 0 , sizeof(sum_r));
        cin >> n ;
        char p ;
        for( int i = 1 ; i <= n ; i ++ ){
            cin >> p ;
            if( p == '<' ){
                pre_l[i] ++ ;
                pre_l[i] += pre_l[i-1] ;
                pre_r[i] = pre_r[i-1] ;
            }else{
                pre_r[i] ++ ;
                pre_r[i] += pre_r[i-1] ;
                pre_l[i] = pre_l[i-1] ;
            }
        }
        for( int i = 1 ; i <= n ; i ++ ){
            if( pre_r[i] > pre_r[i-1] ){
                sum_r[pr] += i ;
                sum_r[pr] += sum_r[pr-1] ;
                pr ++ ;
            }else{
                sum_r[pr] = sum_r[pr-1] ;
                pr ++ ;
            }
            if( pre_l[i] > pre_l[i-1] ){
                sum_l[pl] += i ;
                sum_l[pl] += sum_l[pl-1] ;
                pl ++ ;
            }else{
                sum_l[pl] = sum_l[pl-1] ;
                pl ++ ;
            }
        }
        for( int i = 1 ; i <= n ; i ++ ){
            int ans = 0 ;
            //如果这里是向左的
            if( pre_l[i] > pre_l[i-1] ){
                if( pre_r[i] > ( pre_l[n] - pre_l[i+1])){
                    //若最后是从右侧跑出 
                    ans = (i + n + 1 + 2 * ( sum_l[pl] - sum_r[pl+1])) ;
                }else{
                    //若最后是从左侧跑出 
                    ans = (i + 2 * (sum_l[pl] - sum_r[pl])) ;
                }
            }
            //如果这里是向右的
            if( pre_l[i] == pre_l[i-1] ){
                if(pre_r[i-1] > ( pre_l[n] - pre_l[i])){
                    //若最后从右侧跑出 
                    ans =  (2 * ( sum_l[pl] - sum_r[pl])  )+ n + 1 - i ;
                }else{
                    //若最后从左侧跑出 
                    ans = ( 2 * ( sum_l[pr+1] - sum_r[pr])) - i; 
                }
            }
            printf("%d " , ans );
        }
        printf("\n");
        for( int i = 1 ; i <= pl ; i ++ ){
            cout << sum_l[i] << " ";
        }
        cout << endl;
        for( int i = 1 ; i <= pr ; i ++ ){
            cout << sum_r[i] << " ";
        }
        cout  <<endl;
        for( int i = 1 ; i <= n ; i ++ ){
            cout << pre_l[i] << " " ;
        }
        cout << endl;
        for ( int i = 1  ; i <= n ; i ++ ){
        cout << pre_r[i] << " " ;
        }
        cout << endl << endl;
    }
    return 0 ;
}

感谢大佬！好人一生平安！