제출 #689474

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
689474		coding_snorlax	DNA 돌연변이 (IOI21_dna)	C++17	100 / 100	35 ms	5428 KiB

dna

#include<bits/stdc++.h>
#include "dna.h"
using namespace std;
#define pb push_back
vector<int> A_T={0},T_A={0},A_C={0},C_A={0},C_T={0},T_C={0};
void init(string a, string b){
    for(int i=0;i<(int)a.size();i++){
        if(a[i]=='A' && b[i]=='T'){
            A_T.pb(A_T.back()+1);T_A.pb(T_A.back());A_C.pb(A_C.back());
            C_A.pb(C_A.back());C_T.pb(C_T.back());T_C.pb(T_C.back());
        }
        else if(a[i]=='T' && b[i]=='A'){
            A_T.pb(A_T.back());T_A.pb(T_A.back()+1);A_C.pb(A_C.back());
            C_A.pb(C_A.back());C_T.pb(C_T.back());T_C.pb(T_C.back());
        }
        else if(a[i]=='A' && b[i]=='C'){
            A_T.pb(A_T.back());T_A.pb(T_A.back());A_C.pb(A_C.back()+1);
            C_A.pb(C_A.back());C_T.pb(C_T.back());T_C.pb(T_C.back());
        }
        else if(a[i]=='C' && b[i]=='A'){
            A_T.pb(A_T.back());T_A.pb(T_A.back());A_C.pb(A_C.back());
            C_A.pb(C_A.back()+1);C_T.pb(C_T.back());T_C.pb(T_C.back());
        }
        else if(a[i]=='C' && b[i]=='T'){
            A_T.pb(A_T.back());T_A.pb(T_A.back());A_C.pb(A_C.back());
            C_A.pb(C_A.back());C_T.pb(C_T.back()+1);T_C.pb(T_C.back());
        }
        else if(a[i]=='T' && b[i]=='C'){
            A_T.pb(A_T.back());T_A.pb(T_A.back());A_C.pb(A_C.back());
            C_A.pb(C_A.back());C_T.pb(C_T.back());T_C.pb(T_C.back()+1);
        }
        else{
            A_T.pb(A_T.back());T_A.pb(T_A.back());A_C.pb(A_C.back());
            C_A.pb(C_A.back());C_T.pb(C_T.back());T_C.pb(T_C.back());

        }
        //cout<<" "<<A_T.size()<<" "<<i<<"\n";
    }
    //cout<<A_T.size()<<" ";
}
int get_distance(int x,int y){
    int a = A_T[y+1]-A_T[x] , b = T_A[y+1]-T_A[x] , c = A_C[y+1]-A_C[x], d= C_A[y+1]-C_A[x],e = C_T[y+1]-C_T[x],f=T_C[y+1]-T_C[x];
    //cout << a << " "<<b<<" "<< c<<" "<<d<<" "<<" "<<e<<" "<<f;
    if(a+c != b+d || b+f != a+e || d+e != c+f) return -1;
    else{
        int Count = min(a,b)+min(c,d)+min(e,f);
        return Count + (a+b+c+d+e+f-2*Count)*2 /3;
    }
}

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