Submission #789039

#	Time	Username	Problem	Language	Result	Execution time	Memory
789039		Mouad_ouj	Mutating DNA (IOI21_dna)	C++17	100 / 100	49 ms	8584 KiB

dna

#include<bits/stdc++.h>
//#include"prize.h"
using namespace std;
vector<int> ta,tc,ac,at,ca,ct;
vector<int> pret1,prea1,prec1,pret2,prea2,prec2;
//t a,c     a t,c      c  a,t
void init(string a, string b)
{
    int n=a.size()+1;
    pret1.resize(n),prea1.resize(n),prec1.resize(n),pret2.resize(n),prea2.resize(n),prec2.resize(n);
    ta.resize(n),tc.resize(n),ac.resize(n),at.resize(n),ca.resize(n),ct.resize(n);
    pret1[0]=0,pret2[0]=0,prea1[0]=0,prea2[0]=0,prec1[0]=0,prec2[0]=0;
    ta[0]=0,tc[0]=0,ca[0]=0,ct[0]=0,ac[0]=0,at[0]=0;
    for(int x=1;x<n;x++)
    {
        pret1[x]=pret1[x-1],prea1[x]=prea1[x-1],prec1[x]=prec1[x-1];
        if(a[x-1]=='T')
        pret1[x]++;
        if(a[x-1]=='C')
        prec1[x]++;
        if(a[x-1]=='A')
        prea1[x]++;
        pret2[x]=pret2[x-1],prea2[x]=prea2[x-1],prec2[x]=prec2[x-1];
        if(b[x-1]=='T')
        pret2[x]++;
        if(b[x-1]=='C')
        prec2[x]++;
        if(b[x-1]=='A')
        prea2[x]++;
        ta[x]=ta[x-1],tc[x]=tc[x-1];
        at[x]=at[x-1],ac[x]=ac[x-1];
        ca[x]=ca[x-1],ct[x]=ct[x-1];
        if(a[x-1]=='A' && b[x-1]=='C')
            ac[x]++;
        if(a[x-1]=='A' && b[x-1]=='T')
            at[x]++;
        if(a[x-1]=='T' && b[x-1]=='C')
            tc[x]++; 
        if(a[x-1]=='T' && b[x-1]=='A')
            ta[x]++;
        if(a[x-1]=='C' && b[x-1]=='A')
            ca[x]++;
        if(a[x-1]=='C' && b[x-1]=='T')
            ct[x]++;
    }
}
int get_distance(int y, int x)
{
    x++,y++;
    if(prea1[x]-prea1[y-1]==prea2[x]-prea2[y-1] && pret1[x]-pret1[y-1]==pret2[x]-pret2[y-1] && prec1[x]-prec1[y-1]==prec2[x]-prec2[y-1])
    {
        int ans=0,cur=0;
        ans=min(ac[x]-ac[y-1],ca[x]-ca[y-1])+min(at[x]-at[y-1],ta[x]-ta[y-1])+min(ct[x]-ct[y-1],tc[x]-tc[y-1]);
        cur=max(ac[x]-ac[y-1],ca[x]-ca[y-1])+max(at[x]-at[y-1],ta[x]-ta[y-1])+max(ct[x]-ct[y-1],tc[x]-tc[y-1])-ans;
        ans+=(cur/3)*2;
        return ans;
    }
    else
    return -1;
}

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