提出 #577608

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ソースコード 拡げる

#include<vector>
#include<iostream>
#include<map>
#include<numeric>
#include<algorithm>
#include<set>
#include<queue>
#include<cassert>
#include<cstring>
#include<cstdio>
#include<cstdlib>

using namespace std;
#define REP(i, n) for(int i=0;i<(int)n;i++)
#define REPS(i, n) for(int i=1;i<=(int)n;i++)
#define RREP(i, n) for(int i=n-1;i>=0;i--)
#define RREPS(i, n) for(int i=n;i>0;i--)
#define ALL(v) v.begin(), v.end()
#define RALL(v) v.rbegin(), v.rend()
#define pb push_back
#define eb emplace_back
#define mp make_pair

typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

template<class T>
ostream &operator<<(ostream &os, const vector<T> &v){
	os << "[";
	REP(i, v.size()){
		if(i) os << ",";
		os << v[i];
	}
	return os << "]";
}

const int INF = 1<<29;	
int T, n, m;
string s;

vi mul(const vi &a, const vi &b){
	vi res(a.size());
	REP(i, a.size()) res[i] = b[a[i]];
	return res;
}

vi pow(vi a, ll k){
	vi res(n*n);
	iota(ALL(res), 0);
	while(k){
		if(k&1) res = mul(res, a);
		a = mul(a, a);
		k /= 2;
	}
	return res;
}

ll num(int &i){
	ll res = 0;
	while(i < s.size() && isdigit(s[i])){
		res = res * 10 + s[i ++] - '0';
	}
	return res;
}

vi ope(int &i){
	char c = s[i++];
	int k = num(i) - 1;
	vi res(n*n);
	iota(ALL(res), 0);
	REP(i, n){
		if(c == 'D') res[i*n+k] = (i+1)%n*n+k;
		else if(c == 'U') res[i*n+k] = (i+n-1)%n*n+k;
		else if(c == 'R') res[k*n+i] = k*n+(i+1)%n;
		else if(c == 'L') res[k*n+i] = k*n+(i+n-1)%n;
		else assert(false);
	}
	return res;
}

vi seq(int &i){
	vi res(n*n);
	iota(ALL(res), 0);
	if(i == s.size()) return res;
	while(i < s.size() && s[i] != ')'){
		vi ret;
		if(s[i] == '('){
			ret = seq(++i);
			assert(s[i++] == ')');
			ll k = num(i);
			ret = pow(ret, k);
		}else if(isupper(s[i])){
			ret = ope(i);
		}else assert(false);
		res = mul(res, ret);
	}
	return res;
}

int main(){
	while(cin >> n >> m >> s){
		int i = 0;
		vi res = seq(i);

		vi ans(n*n);
		REP(i, n*n) ans[res[i]] = i + 1;
		REP(i, n)REP(j, n){
			cout << ans[i*n+j] << " \n"[j+1 == n];
		}
	}
	return 0;	
}

提出情報

ジャッジ結果