题解

简单小链表
幸好数据水，不然应该过不了

//0+0=-1
//0+1=1
//朝向:!xor(turn, tmp)
#include <iostream>
#include <cstdio>
using namespace std;
struct node{
	node *pre = nullptr, *next = nullptr;
	string name;
	bool turn;
};
int main(){
	node *head = new node;
	node *r = head;
	int n, t;
	cin >> n >> t;
	for(int i = 1; i <= n; i++){
		node *p = new node;
		p -> pre = r, r -> next = p;
		r = p;
		cin >> p -> turn >> p -> name;
	}
	r -> next = head -> next, head -> next -> pre = r; 
	node *p = head -> next;
	bool tmp;
	int m;
	while(t--){
		cin >> tmp >> m;
		if(tmp == p -> turn){
			for(int i = 1; i <= m; i++){
				p = p -> pre;
			}
		}else{
			for(int i = 1; i <= m; i++){
				p = p -> next;
			}
		}
	}
	cout << p -> name;

	return 0;
}

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没想到吧这才是真正的题解

#include <iostream>
#include <cstdio>
using namespace std;
struct node{
	string name;
	bool turn;
}a[100005];
int main(){
	int n, t;
	cin >> n >> t;
	for(int i = 1; i <= n; i++){
		cin >> a[i].turn >> a[i].name;
	}
	int tmp, m, cur = 1;
	while(t--){
		cin >> tmp >> m;
		tmp = (a[cur].turn ^ tmp) ^ 0;
		if(tmp == 0) tmp--;
		cur += tmp * m;
		if(cur < 1) cur += n;
		if(cur > n) cur -= n;
	}
	cout << a[cur].name;

	return 0;
}