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;
#define scd(t) scanf("%d", &t)
#define sclld(t) scanf("%lld", &t)
#define forr(i, j, k) for (int i = j; i < k; i++)
#define frange(i, j) forr(i, 0, j)
#define all(cont) cont.begin(), cont.end()
#define mp make_pair
#define pb push_back
#define f first
#define s second
typedef long long int lli;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<bool> vb;
typedef vector<lli> vll;
typedef vector<string> vs;
typedef vector<pii> vii;
typedef vector<vi> vvi;
typedef map<int, int> mpii;
typedef set<int> seti;
typedef multiset<int> mseti;
typedef long double ld;
vi cnta1, cntc1, cntt1;
vi cnta2, cntc2, cntt2;
vi inc;
void init(std::string a, std::string b) {
int n = a.size();
cnta1 = cntc1 = cntt1 = vi(n+1);
cnta2 = cntc2 = cntt2 = vi(n+1);
inc = vi(n+1);
forr(i, 1, n+1) {
cnta1[i] = (a[i-1] == 'A') + cnta1[i-1];
cntc1[i] = (a[i-1] == 'C') + cntc1[i-1];
cntt1[i] = (a[i-1] == 'T') + cntt1[i-1];
cnta2[i] = (b[i-1] == 'A') + cnta2[i-1];
cntc2[i] = (b[i-1] == 'C') + cntc2[i-1];
cntt2[i] = (b[i-1] == 'T') + cntt2[i-1];
inc[i] = inc[i-1] + (a[i-1] != b[i-1]);
}
}
int get_distance(int x, int y) {
x++;
y++;
// printf("%d %d\n", cnta1[y] - cnta1[x-1], cnta2[y] - cnta2[x-1]);
if(cnta1[y] - cnta1[x-1] != cnta2[y] - cnta2[x-1]) return -1;
if(cntc1[y] - cntc1[x-1] != cntc2[y] - cntc2[x-1]) return -1;
if(cntt1[y] - cntt1[x-1] != cntt2[y] - cntt2[x-1]) return -1;
return max(0, inc[y] - inc[x-1] - 1);
}
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