Submission #548724

#	Time	Username	Problem	Language	Result	Execution time	Memory
548724		esomer	Mutating DNA (IOI21_dna)	C++17	56 / 100	37 ms	5768 KiB

dna

#include<bits/stdc++.h>
#include "dna.h"

using namespace std;

vector<tuple<int, int, int>> prefix1;
vector<tuple<int, int, int>> prefix2;
vector<int> prefix;
int n;

void init(string a, string b) {
	n = a.size();
	prefix.resize(n);
	prefix1.resize(n);
	prefix2.resize(n);
	for(int i = 0; i < n; i++){
		if(i == 0){
			prefix1[i] = {0, 0, 0};
			prefix2[i] = {0, 0, 0};
			if(a[i] == 'A'){
				get<0> (prefix1[i])++;
			}else if(a[i] == 'T'){
				get<1>(prefix1[i])++;
			}else get<2> (prefix1[i])++;
			if(b[i] == 'A'){
				get<0> (prefix2[i])++;
			}else if(b[i] == 'T'){
				get<1>(prefix2[i])++;
			}else get<2> (prefix2[i])++;
		}else{
			prefix1[i] = prefix1[i - 1];
			prefix2[i] = prefix2[i - 1];
			if(a[i] == 'A'){
				get<0> (prefix1[i])++;
			}else if(a[i] == 'T'){
				get<1>(prefix1[i])++;
			}else get<2> (prefix1[i])++;
			if(b[i] == 'A'){
				get<0> (prefix2[i])++;
			}else if(b[i] == 'T'){
				get<1>(prefix2[i])++;
			}else get<2> (prefix2[i])++;
		}
		if(a[i] != b[i]){
			if(i == 0) prefix[i] = 1;
			else prefix[i] = prefix[i-1] + 1;
		}else{
			if(i == 0) prefix[i] = 0;
			else prefix[i] = prefix[i-1];
		}
	}
}

int get_distance(int x, int y) {
	int da, db;
	if(x > 0) da = get<0> (prefix1[y]) - get<0> (prefix1[x-1]);
	else da = get<0> (prefix1[y]);
	if(x > 0) db = get<0> (prefix2[y]) - get<0> (prefix2[x-1]);
	else db = get<0> (prefix2[y]);
	if(da != db) return -1;
	if(x > 0) da = get<1> (prefix1[y]) - get<1> (prefix1[x-1]);
	else da = get<1> (prefix1[y]);
	if(x > 0) db = get<1> (prefix2[y]) - get<1> (prefix2[x-1]);
	else db = get<1> (prefix2[y]);
	if(da != db) return -1;
	if(x > 0) da = get<2> (prefix1[y]) - get<2> (prefix1[x-1]);
	else da = get<2> (prefix1[y]);
	if(x > 0) db = get<2> (prefix2[y]) - get<2> (prefix2[x-1]);
	else db = get<2> (prefix2[y]);
	if(da != db) return -1;

	int d;
	if(x > 0) d = prefix[y] - prefix[x-1];
	else d = prefix[y];
	if(d % 2 != 0) return d / 2 + 1;
	else return d / 2;
}

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