Java(javac 1.8) 解法, 执行用时: 299ms, 内存消耗: 24584K, 提交时间: 2019-07-22 16:30:05

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String S = sc.next();
		int r = 0, s = 0;
        long ans = 0;
		for (int i = 0; i < S.length(); i++) {
			if (S.charAt(i) == 'r')
				r = i + 1;
			if (S.charAt(i) == 'x')
				s = i + 1;
			ans = ans + Math.min(r, s);
			
		}
		System.out.println(ans);
	}

}

C++11(clang++ 3.9) 解法, 执行用时: 52ms, 内存消耗: 3604K, 提交时间: 2019-06-08 22:21:26

#include<bits/stdc++.h>
using namespace std;
string s;
int main()
{
    long long ans=0,ri=-1,xi=-1;
    cin>>s;
    for(int i=0;i<s.size();++i)
    {
        if(s[i]=='x') xi=i;
        else if(s[i]=='r') ri=i;
        ans+=min(ri,xi)+1;
    }
    cout<<ans<<endl;
    return 0;
}

C++14(g++5.4) 解法, 执行用时: 51ms, 内存消耗: 3616K, 提交时间: 2019-07-07 09:44:25

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
int main()
{
	string s;cin>>s;	
	int rr=0,xx=0;
	ll ans=0;
	for(int i=0;s[i];i++)
	{
		if(s[i]=='r') rr = i+1;
		if(s[i]=='x') xx = i+1;
		ans+=min(rr,xx);
	}
	printf("%lld",ans);
}