이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "dna.h"
#include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
string ga, gb;
map<char, vi> prefixCharA, prefixCharB;
// prefix array of the positions where [c1][c2] c2 is in b instead of c1
map<char, map<char, int>> inPlaceOf[(int)1e5 + 1];
char characters[3] = {'A', 'T', 'C'};
void init(string a, string b)
{
int n = a.size();
ga = ' ' + a;
gb = ' ' + b;
for (char c : characters)
{
prefixCharA[c] = prefixCharB[c] = vi(n + 1, 0);
for (int i = 1; i <= n; i++)
{
prefixCharA[c][i] = (ga[i] == c) + prefixCharA[c][i - 1];
prefixCharB[c][i] = (gb[i] == c) + prefixCharB[c][i - 1];
}
}
for (int i = 1; i <= n; i++)
{
for (int j = 0; j < 3; j++)
for (int k = 0; k < 3; k++)
{
if (k == j)
continue;
inPlaceOf[i][characters[j]][characters[k]] =
inPlaceOf[i - 1][characters[j]][characters[k]];
}
if (a[i] != b[i])
inPlaceOf[i][a[i]][b[i]]++;
}
}
int get_distance(int x, int y)
{
x++, y++;
bool pos = true;
for (char c : characters)
{
int countInA = prefixCharA[c][y] - prefixCharA[c][x - 1];
int countInB = prefixCharB[c][y] - prefixCharB[c][x - 1];
pos &= (countInA == countInB);
}
if (!pos)
return -1;
int moves = 0;
moves = inPlaceOf[y]['A']['T'] - inPlaceOf[x - 1]['A']['T'];
return moves / 2;
}
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