思路

关键点在任意时间内都通过率大于等于50%的题为签到题这句话
每一次都进行判断是不是满足条件即可

代码

// Problem: 九峰与签到题
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/9984/A
// Memory Limit: 524288 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp(aa,bb) make_pair(aa,bb)
#define _for(i,b) for(int i=(0);i<(b);i++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,b,a) for(int i=(b);i>=(a);i--)
#define mst(abc,bca) memset(abc,bca,sizeof abc)
#define X first
#define Y second
#define lowbit(a) (a&(-a))
#define debug(a) cout<<#a<<":"<<a<<"\n"
typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
typedef long double ld;
const int N=100010;
const int INF=0x3f3f3f3f;
const int mod=1e9+7;
const double eps=1e-6;
const double PI=acos(-1.0);

int m,n;
int sum[N],ac[N];
vector<int> ans;
bool st[N];

void check(){
    rep(id,1,n){
        if(sum[id]){
            if(2*ac[id]<sum[id]) st[id]=0;
        }
    }
}

void solve(){
    cin>>m>>n;
    rep(id,1,n) st[id]=1;
    while(m--){
        int id;string op;
        cin>>id>>op;
        sum[id]++;
        if(op=="AC") ac[id]++;
        check();
    }
    rep(id,1,n){
        if(sum[id]){
            if(st[id]) ans.pb(id);
        }
    }

    if(!ans.size()) cout<<"-1\n";
    else {
        for(int x:ans) cout<<x<<" ";
    }
}


int main(){
    ios::sync_with_stdio(0);cin.tie(0);
//    int t;cin>>t;while(t--)
    solve();
    return 0;
}