Submission #1142069

#	Time	Username	Problem	Language	Result	Execution time	Memory
1142069		vyaduct	Mutating DNA (IOI21_dna)	C++20	56 / 100	31 ms	5704 KiB

dna

#include <bits/stdc++.h>
#include "dna.h"
#define sz(a) (int)(a).size()
using namespace std;

const int N = 100'001;
int cntA[N][3], cntB[N][3];
int pref[N];
string A, B;

int sum(int ch, int x, int y, bool A){
  if (A) return cntA[y][ch]-(x == 0 ? 0 : cntA[x-1][ch]);
  else return cntB[y][ch]-(x == 0 ? 0 : cntB[x-1][ch]);
}

int ord(char c){
  // index by char
  if (c == 'A') return 0;
  else if (c == 'T') return 1;
  else return 2;
}

bool check(int x, int y){
  for (int i=0;i<3;i++){
    int sumA = sum(i, x, y, true);
    int sumB = sum(i, x, y, false);
    if (sumA != sumB){
      return false;
    }
  }
  return true;
}

void init(string a, string b) {
  A = a, B = b;
  memset(cntA, 0, sizeof(cntA));
  memset(cntB, 0, sizeof(cntB));
  // prefsum precomputation
  int n = sz(a);
  for (int i=0;i<n;i++){
    // string a
    cntA[i][0] = (i == 0 ? 0 : cntA[i-1][0]) + (ord(a[i]) == 0);
    cntA[i][1] = (i == 0 ? 0 : cntA[i-1][1]) + (ord(a[i]) == 1);
    cntA[i][2] = (i == 0 ? 0 : cntA[i-1][2]) + (ord(a[i]) == 2);
    // string b
    cntB[i][0] = (i == 0 ? 0 : cntB[i-1][0]) + (ord(b[i]) == 0);
    cntB[i][1] = (i == 0 ? 0 : cntB[i-1][1]) + (ord(b[i]) == 1);
    cntB[i][2] = (i == 0 ? 0 : cntB[i-1][2]) + (ord(b[i]) == 2);
  }

  for (int i=0;i<n;i++){
    pref[i] = (i == 0 ? 0 : pref[i-1]) + (A[i] == 'T' && B[i] == 'A');
  }
}
// Accepted first subtask, (y-x <= 2)
int subtask1(int x, int y){
  int size = y-x+1;
  if (size == 1) return 0;
  if (size == 2){
    return A[x] == B[x] ? 0 : 1;
  } else {
    int diff=0;
    for (int i=x;i<=y;i++){
      diff += (A[i] == B[i] ? 0 : 1);
    }
    if (diff != 0) return diff-1;
    else return diff;
  }
}

int subtask2_3(int x, int y){
  int res = pref[y]-(x==0 ? 0 : pref[x-1]);
  return res;
}

int get_distance(int x, int y) {
  bool possible = check(x,y);
  if (!possible){
    return -1;
  }
  if (y-x <= 2) return subtask1(x,y);
  else return subtask2_3(x,y);
}

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