2.1 KB | 2184 chars
/*
[2840: 행운의 바퀴](https://www.acmicpc.net/problem/2840)
Tier: Silver 4
Category: implementation, simulation
*/
#include <bits/stdc++.h>
using namespace std;
#define for1(s, e) for(int i = s; i < e; i++)
#define for1j(s, e) for(int j = s; j < e; j++)
#define forEach(k) for(auto i : k)
#define forEachj(k) for(auto j : k)
#define sz(vct) vct.size()
#define all(vct) vct.begin(), vct.end()
#define sortv(vct) sort(vct.begin(), vct.end())
#define uniq(vct) sort(all(vct));vct.erase(unique(all(vct)), vct.end())
#define fi first
#define se second
#define INF (1ll << 60ll)
typedef unsigned long long ull;
typedef long long ll;
typedef ll llint;
typedef unsigned int uint;
typedef unsigned long long int ull;
typedef ull ullint;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;
typedef pair<double, int> pdi;
typedef pair<string, string> pss;
typedef vector<int> iv1;
typedef vector<iv1> iv2;
typedef vector<ll> llv1;
typedef vector<llv1> llv2;
typedef vector<pii> piiv1;
typedef vector<piiv1> piiv2;
typedef vector<pll> pllv1;
typedef vector<pllv1> pllv2;
typedef vector<pdd> pddv1;
typedef vector<pddv1> pddv2;
const double EPS = 1e-8;
const double PI = acos(-1);
template<typename T>
T sq(T x) { return x * x; }
int sign(ll x) { return x < 0 ? -1 : x > 0 ? 1 : 0; }
int sign(int x) { return x < 0 ? -1 : x > 0 ? 1 : 0; }
int sign(double x) { return abs(x) < EPS ? 0 : x < 0 ? -1 : 1; }
ll n, k;
vector <char> ar;
void solve() {
cin >> n >> k;
ar.resize(n, '0');
ll crt = 0;
for1(0, k) {
ll num;
char a;
cin >> num >> a;
crt -= num;
crt += n * 100000;
crt %= n;
if(ar[crt] == '0') ar[crt] = a;
else if(ar[crt] != a) {
cout << "!";
return;
}
}
for1(0, n) {
for1j(i + 1, n) {
if(ar[i] != '0' && ar[i] == ar[j]) {
cout << "!";
return;
}
}
}
for1(crt, n) {
if(ar[i] == '0') cout << '?';
else cout << ar[i];
}
for1(0, crt) {
if(ar[i] == '0') cout << '?';
else cout << ar[i];
}
}
int main() {
ios::sync_with_stdio(0);
cin.tie(NULL);cout.tie(NULL);
int tc = 1; // cin >> tc;
while(tc--) solve();
}