Submission #708539

#	Time	Username	Problem	Language	Result	Execution time	Memory
708539		ngano_upat_na	Mutating DNA (IOI21_dna)	C++17	35 / 100	42 ms	7628 KiB

dna

#include "dna.h"
#include "bits/stdc++.h"
using namespace std;
using ll = long long;

const int N = 100001;

vector<int> A0(N,0), A1(N,0);
vector<int> T0(N,0), T1(N,0);
vector<int> C0(N,0), C1(N,0);
vector<int> AC(N,0), CA(N,0), AT(N,0), TA(N,0), TC(N,0), CT(N,0);

string s, t;

void init(string a, string b) {
    s = a;
    t = b;
    int sz = (int)a.size();
    for (int i=0; i<sz; i++) {
        A0[i+1] = A0[i];
        T0[i+1] = T0[i];
        C0[i+1] = C0[i];
        
        if (s[i] == 'A') A0[i+1]++;
        if (s[i] == 'T') T0[i+1]++;
        if (s[i] == 'C') C0[i+1]++;
    }
    for (int i=0; i<sz; i++) {
        A1[i+1] = A1[i];
        T1[i+1] = T1[i];
        C1[i+1] = C1[i];
        
        if (t[i] == 'A') A1[i+1]++;
        if (t[i] == 'T') T1[i+1]++;
        if (t[i] == 'C') C1[i+1]++;
    } 
    for (int i=0; i<sz; i++) {
        AC[i+1] = AC[i];
        CA[i+1] = CA[i];
        AT[i+1] = AT[i];
        TA[i+1] = TA[i];
        CT[i+1] = CT[i];
        TC[i+1] = CT[i];
        
        if (s[i] == 'A' && t[i] == 'C') AC[i+1]++;
        if (s[i] == 'C' && t[i] == 'A') CA[i+1]++;
        if (s[i] == 'A' && t[i] == 'T') AT[i+1]++;
        if (s[i] == 'T' && t[i] == 'A') TA[i+1]++;
        if (s[i] == 'C' && t[i] == 'T') CT[i+1]++;
        if (s[i] == 'T' && t[i] == 'C') TC[i+1]++;
    }
}   

int get_distance(int x, int y) {
    int ans = 0;
    y++;
    if ((A0[y] - A0[x] != A1[y] - A1[x]) || (T0[y] - T0[x] != T1[y] - T1[x]) || (C0[y] - C0[x] != C1[y] - C1[x])) return -1;
    
    int ac = AC[y] - AC[x], ca = CA[y] - CA[x];
    int at = AT[y] - AT[x], ta = TA[y] - TA[x];
    int ct = CT[y] - CT[x], tc = TC[y] - TC[x];
    
    int a = min(ac,ca);
    int b = min(at,ta);
    int c = min(ct,tc);
    ans = ans + a + b + c;
    ac -= a;
    ca -= a;
    at -= b;
    ta -= b;
    ct -= c;
    tc -= c;
    
    int d = min(ac,min(ta,ct));
    ans += (2*d);
    ac -= d;
    ta -= d;
    ct -= d;
    
    int e = min(ca,min(at,tc));
    ans += (2*e);
    ca -= e;
    at -= e;
    tc -= e;
    
    int f = min(at,min(tc,ca));
    ans += (2*f);
    at -= f;
    tc -= f;
    ca -= f;
    
    int g = min(ta,min(ct,ac));
    ans += (2*g);
    ta -= g;
    ct -= g;
    ac -= g;
    
    if (ac == 0 && ca == 0 && ct == 0 && tc == 0 && at == 0 && ta == 0) {
        return ans;
    }
    else {
        return -1;
    }   
}   

• Subtask 1: 0 / 21
• Subtask 2: 22 / 22
• Subtask 3: 13 / 13
• Subtask 4: 0 / 28
• Subtask 5: 0 / 16