Submission #1193152

#	Time	Username	Problem	Language	Result	Execution time	Memory
1193152		trvhung	DNA 돌연변이 (IOI21_dna)	C++20	100 / 100	26 ms	8780 KiB

dna

#include "dna.h"
#include <bits/stdc++.h>
// #include <ext/rope>
// #include <ext/pb_ds/assoc_container.hpp>

// using namespace __gnu_pbds;
// using namespace __gnu_cxx;
using namespace std;

// #define   ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define            ll long long
#define           ull unsigned long long
#define            ld long double
#define            pb push_back
#define  bit(mask, i) ((mask >> i) & 1)
#define            el '\n'
#define             F first
#define             S second

template <class X, class Y> bool maximize(X &x, const Y &y) { return (x < y ? x = y, 1 : 0); }
template <class X, class Y> bool minimize(X &x, const Y &y) { return (x > y ? x = y, 1 : 0); }

const int INF = 1e9;
const ll LINF = 1e18;
const int MOD = 1e9 + 7;
const int MULTI = 0;
const ld eps = 1e-9;
const int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0}; // R D L U
const int ddx[4] = {-1, 1, 1, -1}, ddy[4] = {1, 1, -1, -1}; // UR DR DL UL
const char cx[4] = {'R', 'D', 'L', 'U'};
const ll base = 31;
const int nMOD = 2;
const ll mods[] = {(ll)1e9 + 10777, (ll)1e9 + 19777, (ll)1e9 + 3, (ll)1e9 + 3777};

const int MAX = 1e5 + 5;
const char ch[3] = {'A', 'C', 'T'};
int n, prefChar[2][3][MAX], prefPairs[3][3][MAX];
string s, t;

void init(string a, string b) {
	n = (int) a.size();
	s = ' ' + a; t = ' ' + b;

	for (int i = 1; i <= n; ++i)
		for (int j = 0; j < 3; ++j) {
			prefChar[0][j][i] = prefChar[0][j][i - 1] + (s[i] == ch[j]);
			prefChar[1][j][i] = prefChar[1][j][i - 1] + (t[i] == ch[j]);
		}

	for (int i = 1; i <= n; ++i)
		for (int j = 0; j < 3; ++j)
			for (int k = 0; k < 3; ++k)
				prefPairs[j][k][i] = prefPairs[j][k][i - 1] + (s[i] == ch[j] && t[i] == ch[k]);
}

int get_distance(int x, int y) {
	++x; ++y;

	for (int j = 0; j < 3; ++j)
		if (prefChar[0][j][y] - prefChar[0][j][x - 1] != prefChar[1][j][y] - prefChar[1][j][x - 1])
			return -1;

	int res = 0, rem = 0;
	for (int i = 0; i < 2; ++i)
		for (int j = i + 1; j < 3; ++j) {
			int cntIJ = prefPairs[i][j][y] - prefPairs[i][j][x - 1];
			int cntJI = prefPairs[j][i][y] - prefPairs[j][i][x - 1];
			res += min(cntIJ, cntJI);
			rem += max(cntIJ, cntJI) - min(cntIJ, cntJI);
		}

	res += 2 * (rem / 3);

	return res;
}

• Subtask 1: 21 / 21
• Subtask 2: 22 / 22
• Subtask 3: 13 / 13
• Subtask 4: 28 / 28
• Subtask 5: 16 / 16