玩具谜题

https://www.luogu.org/problemnew/show/P1563

题解：题目不难，主要第一次遇到异或，就想写一下

C++版本一

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
using namespace std;
typedef long long ll;
typedef __int128 lll;
const int N=100000+1000;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,q;
map< int,string>s;
bool a[N];
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    char str[20];

    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++){
        scanf("%d %s",&k,str);
        a[i]=k;
        s[i]=str;
    }
    int pos=1;
    while(m--){
        scanf("%d %d",&k,&q);
        if(a[pos]^k){
            pos+=q;
        }else{
            pos-=q;
        }
        if(pos<=0){
            pos+=n;
        }
        if(pos>n){
            pos-=n;
        }
    }
    cout << s[pos] << endl;
    //cout << "Hello world!" << endl;
    return 0;
}

C++版本二

include<bits/stdc++.h>
using namespace std;
struct node 
{
    int head;
    string name;
}a[100005];
int n,m,x,y;
int main()
{
    cin>>n>>m;
    for(int i=0;i<n;i++)
    {
        cin>>a[i].head>>a[i].name;
    }
    int now=0;
    for(int i=1;i<=m;i++)
    {
        cin>>x>>y;
        if(a[now].head==0&&x==0)now=(now+n-y)%n;
        else if(a[now].head==0&&x==1)now=(now+y)%n;
        else if(a[now].head==1&&x==0)now=(now+y)%n;
        else if(a[now].head==1&&x==1)now=(now+n-y)%n;
    }
    cout<<a[now].name<<endl;
    return 0;
}

JAVA版本一

import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;

public class Main {
    static int place, n, m;// place是当前位置,n是人物个数,m是指令个数
    static int[] N;// 存储人物朝向的数组
    static String[] Name;// 存储人物名称的数组

    public static void main(String[] args) {
        InputReader in = new InputReader(System.in);
        n = in.nextInt();
        m = in.nextInt();
        N = new int[n];
        Name = new String[n];
        for (int i = 0; i < n; i++) {// 读取所有的人物
            N[i] = in.nextInt();
            Name[i] = in.next();
        }
        for (int i = 0; i < m; i++) {// 读取所有操作同时计算
            opeart(in.nextInt(), in.nextInt());
        }
        System.out.println(Name[place]);
    }

    static void opeart(int i, int j) {
        // 其实只有两种情况,
        // 1 如果朝向内侧又向左移动表示为 0 0,如果朝向外侧有向右移动表示为1 1,这种情况下应该用当前位置减去移动的距离
        // 2朝向和移动方向不想同0 1或者1 0,则应该用当前位置加上移动的距离
        if (N[place] == i)
            place = (place + n - j) % n;
        else
            place = (place + j) % n;
    }

    // 这是个内部类,借鉴cf上某个java大佬的写法,只是为了解决读取输入太慢的问题
    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

    }
}

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