Submission #1189602

#	Time	Username	Problem	Language	Result	Execution time	Memory
1189602		Panda50O	Mutating DNA (IOI21_dna)	C++20	100 / 100	25 ms	8456 KiB

dna

// #include<iostream>
#include "dna.h"
using namespace std;

const int mxN = 1e5+5;
int qs_a[mxN][3];
int qs_b[mxN][3];
int pp[mxN][3][3];

bool can(int x, int y) {
    if((qs_a[y][0] - qs_a[x-1][0]) != (qs_b[y][0] - qs_b[x-1][0])) return false;
    if((qs_a[y][1] - qs_a[x-1][1]) != (qs_b[y][1] - qs_b[x-1][1])) return false;
    if((qs_a[y][2] - qs_a[x-1][2]) != (qs_b[y][2] - qs_b[x-1][2])) return false;
    return true;
}

int parse(char c) {
    if(c == 'A') return 0;
    if(c == 'C') return 1;
    return 2;
}

void reset(int i) {
    for(int j = 0; j < 3; ++j) {
        qs_a[i][j] = qs_a[i-1][j];
        qs_b[i][j] = qs_b[i-1][j];
    }
    for(int j = 0; j < 3; ++j) {
        for(int k = 0; k < 3; ++k) {
            pp[i][j][k] = pp[i-1][j][k];
        }
    }
    return;
}

void init(string a, string b) {
    int n = a.size();
    for(int i = 1; i <= n; ++i) {
        reset(i);
        

        if(parse(a[i-1]) == parse(b[i-1])) continue;
        qs_a[i][parse(a[i-1])] = qs_a[i-1][parse(a[i-1])] + 1;    
        qs_b[i][parse(b[i-1])] = qs_b[i-1][parse(b[i-1])] + 1;
        pp[i][parse(a[i-1])][parse(b[i-1])] = pp[i-1][parse(a[i-1])][parse(b[i-1])] + 1;
        // cout << pp[i][parse(a[i-1])][parse(b[i-1])] << ' ' << parse(a[i-1]) << ' ' << parse(b[i-1]) << '\n';
    }

    

}

int get_distance(int x, int y) {
    ++x, ++y;
    if(!can(x,y)) {
        return -1;
    }
    int cnt = 0;
    for(int i = 0; i < 3; ++i) {
        cnt += qs_a[y][i] - qs_a[x-1][i];
    }
    
    int cnt_pair = 0;

    for(int i = 0; i < 3; ++i) {
        for(int j = i+1; j < 3; ++j) {
            int p1 = pp[y][i][j] - pp[x-1][i][j];
            int p2 = pp[y][j][i] - pp[x-1][j][i];
            if(p1 > 0 && p2 > 0){ // มีคู่ตรงข้ามกัน  
                cnt_pair += min(p1, p2);
                cnt -= min(p1, p2) * 2;
            } 
        }
    }

    // cout << pp[0][1];

    // return cnt_pair;
    return cnt_pair + (cnt/3)*2;
    // return (all - cnt_pair) % 3 * 2;


}

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