Submission #17876

#	Time	Username	Problem	Language	Result	Execution time	Memory
17876		chrome	K-th path (IZhO11_kthpath)	C++98	0 / 100	0 ms	2272 KiB

kthpath

#include <bits/stdc++.h>

using namespace std;

#define ll long long
#define foreach(it, S) for (__typeof (S.begin()) it = S.begin(); it != S.end(); it++)
#define all(x) x.begin(), x.end()
#define endl '\n'
#define _ ios_base :: sync_with_stdio(false); cin.tie(NULL);

#ifdef inputf
	#define fname ""
#else
	#define fname "" // <- Here
#endif

const double eps = 1e-9;
const int MaxN = int(2e5) + 256;
const int MOD = int(1e9) + 7;

template <typename T> inline T gcd(T a, T b) {
	return b ? gcd (b, a % b) : a;
}

inline bool Palindrome(const string& s) {
	return equal(s.begin(), s.end(), s.rbegin());
}

ll c[100][100];
char a[100][100];

int main() { _
	#ifdef lcl
		freopen(fname".in", "r", stdin);
		freopen(fname".out", "w", stdout);
	#endif

	for (int i = 0; i <= 60; ++i) {
		c[i][0] = c[i][i] = 1;
		for (int j = 1; j < i; ++j)                  
			c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
	}
	
	int n, m; cin >> n >> m;

	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= m; ++j)
			cin >> a[i][j];

	ll k; cin >> k;
	
	int x = 1, y = 1;
	
	int len = 0;
	string res = "";
	
	
	while (len < n + m - 2) {
		// cerr << len << endl;
		res += a[x][y];
		vector <pair <char, pair <int, int> > > v;
		if (y + 1 <= m) v.push_back(make_pair(a[x][y + 1], make_pair(x, y + 1)));
		if (x + 1 <= n) v.push_back(make_pair(a[x + 1][y], make_pair(x + 1, y)));
		sort(all(v));
		for (int i = 0; i < (int)v.size(); ++i) {
			int X = v[i].second.first;
			int Y = v[i].second.second;
			char ch = v[i].first;
			ll cnt = c[n - X + m - Y][n - X];
			// cerr << k << " " << cnt << endl;
			if (k <= cnt) {
				// res += ch;
				x = X; y = Y;
				break;
			} else {
				k -= cnt;
			}
		}
		++len;
	}
	
	res += a[x][y];
	cout << res;

	return 0;
}

Compilation message (stderr)

kthpath.cpp: In function 'int main()':
kthpath.cpp:68:9: warning: unused variable 'ch' [-Wunused-variable]
    char ch = v[i].first;
         ^

• Subtask 1: 0 / 100