Submission #1065652

#	Time	Username	Problem	Language	Result	Execution time	Memory
1065652		c2zi6	Mutating DNA (IOI21_dna)	C++17	100 / 100	30 ms	6296 KiB

dna

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
#define ff first
#define ss second
#define pb push_back
#define all(a) (a).begin(), (a).end()
#define replr(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define reprl(i, a, b) for (int i = int(a); i >= int(b); --i)
#define rep(i, n) for (int i = 0; i < int(n); ++i)
#define mkp(a, b) make_pair(a, b)
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<PII> VPI;
typedef vector<VI> VVI;
typedef vector<VVI> VVVI;
typedef vector<VPI> VVPI;
typedef pair<ll, ll> PLL;
typedef vector<ll> VL;
typedef vector<PLL> VPL;
typedef vector<VL> VVL;
typedef vector<VVL> VVVL;
typedef vector<VPL> VVPL;
template<class T> T setmax(T& a, T b) {if (a < b) return a = b; return a;}
template<class T> T setmin(T& a, T b) {if (a < b) return a; return a = b;}
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template<class T>
using indset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#include "dna.h"

VI pref[6];

/*
 * AC
 * CA
 * AT
 * TA
 * CT
 * TC
 */

void init(string a, string b) {
    int n = a.size();
    rep(k, 6) pref[k] = VI(n+1);
    replr(i, 1, n) {
        string cur;
        cur.pb(a[i-1]);
        cur.pb(b[i-1]);
        if (cur == "AC") pref[0][i]++;
        else if (cur == "CA") pref[1][i]++;
        else if (cur == "AT") pref[2][i]++;
        else if (cur == "TA") pref[3][i]++;
        else if (cur == "CT") pref[4][i]++;
        else if (cur == "TC") pref[5][i]++;
        rep(k, 6) pref[k][i] += pref[k][i-1];
    }
}

int get_distance(int l, int r) {
    int cnt[6];
    rep(k, 6) cnt[k] = pref[k][r+1] - pref[k][l];
    int ans = 0;
    int tmp;
    tmp = min(cnt[0], cnt[1]);
    cnt[0] -= tmp;
    cnt[1] -= tmp;
    ans += tmp;
    tmp = min(cnt[2], cnt[3]);
    cnt[2] -= tmp;
    cnt[3] -= tmp;
    ans += tmp;
    tmp = min(cnt[4], cnt[5]);
    cnt[4] -= tmp;
    cnt[5] -= tmp;
    ans += tmp;
    int A, C, T;
    A = cnt[0] + cnt[2];
    C = cnt[1] + cnt[4];
    T = cnt[3] + cnt[5];
    if (A == C && A == T) return ans + 2*A;
	return -1;
}

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