제출 #1290181

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1290181		blackscreen1	DNA 돌연변이 (IOI21_dna)	C++20	100 / 100	28 ms	12048 KiB

dna

#include "dna.h"
#include <bits//stdc++.h>
using namespace std;
#define ll long long
#define iloop(m, h) for (auto i = m; i != h; i += (m < h ? 1 : -1))
#define jloop(m, h) for (auto j = m; j != h; j += (m < h ? 1 : -1))
#define kloop(m, h) for (auto k = m; k != h; k += (m < h ? 1 : -1))
#define lloop(m, h) for (auto l = m; l != h; l += (m < h ? 1 : -1))
#define iloop_(m, h, g) for (auto i = m; i != h; i += g)
#define jloop_(m, h, g) for (auto j = m; j != h; j += g)
#define kloop_(m, h, g) for (auto k = m; k != h; k += g)
#define lloop_(m, h, g) for (auto l = m; l != h; l += g)
#define getchar_unlocked _getchar_nolock // comment before submission
#define pll pair<ll, ll>
#define plll pair<ll, pll>
#define pllll pair<pll, pll>
#define vll vector<ll>
#define qll queue<ll>
#define dll deque<ll>
#define pqll priority_queue<ll>
#define gll greater<ll>
#define INF 1000000000000000
#define MOD1 1000000007
#define MOD2 998244353
#define MOD3 1000000009
ll pfa[3][1000005], pfb[3][1000005], pfc[6][100005], n;
ll ar[6], ans, k, s;
void init(string a, string b) {
	n = a.length();
	iloop(0, n) {
		jloop(0, 3) pfa[j][i+1] = pfa[j][i];
		jloop(0, 3) pfb[j][i+1] = pfb[j][i];
		jloop(0, 6) pfc[j][i+1] = pfc[j][i];
		if (a[i] == 'A') pfa[0][i+1]++;
		if (a[i] == 'C') pfa[1][i+1]++;
		if (a[i] == 'T') pfa[2][i+1]++;
		if (b[i] == 'A') pfb[0][i+1]++;
		if (b[i] == 'C') pfb[1][i+1]++;
		if (b[i] == 'T') pfb[2][i+1]++;
		if (a[i] == 'A' && b[i] == 'C') pfc[0][i+1]++;
		if (a[i] == 'A' && b[i] == 'T') pfc[1][i+1]++;
		if (a[i] == 'C' && b[i] == 'A') pfc[2][i+1]++;
		if (a[i] == 'C' && b[i] == 'T') pfc[3][i+1]++;
		if (a[i] == 'T' && b[i] == 'A') pfc[4][i+1]++;
		if (a[i] == 'T' && b[i] == 'C') pfc[5][i+1]++;
	}
}

int get_distance(int x, int y) {
	s = 0;
	iloop(0, 3) if (pfa[i][y+1] - pfa[i][x] != pfb[i][y+1] - pfb[i][x]) return -1;
	ans = 0;
	iloop(0, 6) {ar[i] = pfc[i][y+1] - pfc[i][x]; s += ar[i];}
	k = min(ar[0], ar[2]);
	ans += k;
	s -= 2*k;
	k = min(ar[1], ar[4]);
	ans += k;
	s -= 2*k;
	k = min(ar[3], ar[5]);
	ans += k;
	s -= 2*k;
	return ans + s/3*2;
}

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