제출 #694469

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
694469		josanneo22	DNA 돌연변이 (IOI21_dna)	C++17	0 / 100	39 ms	3752 KiB

dna

#include<bits/stdc++.h>
#include<iostream>
#include<stdlib.h>
#include<cmath>
#include <algorithm>
#include<numeric>
using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pair<int, int> > vpii;
typedef pair<ll, ll> pll;
typedef vector<pll> vpll;
typedef vector<ll> vll;
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define trav(a,x) for (auto& a: x)
#define fr(i, a, b, s) for (int i=(a); (s)>0?i<(b):i>=(b); i+=(s))
#define fr1(e) fr(i, 0, e, 1)
#define fr2(i, e) fr(i, 0, e, 1)
#define fr3(i, b, e) fr(i, b, e, 1)
#define mp make_pair
#define pb push_back
#define sz(x) int(x.size())
#define all(x) begin(x), end(x)
#define rall(x) rbegin(x), rend(x)
#define in insert
#define yes cout<<"YES\n"
#define no cout<<"NO\n"
int dx[4] = { -1, 0, 1, 0 };
int dy[4] = { 0, 1, 0, -1 };
const int mod = 1e9 + 7;

void xd(string str)
{
	ios_base::sync_with_stdio(0); cin.tie(0);
	if (str != "")
	{
		//freopen((str + ".in").c_str(), "r", stdin);
		//freopen((str + ".out").c_str(), "w", stdout);
	}
}
int add(int a, int b, int mod = 1e9 + 7) { return (((a % mod) + (b % mod)) + mod) % mod; }
int sub(int a, int b, int mod = 1e9 + 7) { return (((a % mod) - (b % mod)) + mod) % mod; }
int mul(int a, int b, int mod = 1e9 + 7) { return (((a % mod) * (b % mod)) + mod) % mod; }
int bin(int a, int b, int mod = 1e9 + 7) { int ans = 1; while (b) { if (b & 1) ans = mul(ans, a, mod); a = mul(a, a, mod); b >>= 1; }return ans; }
int inverse(int a, int mod = 1e9 + 7) { return bin(a, mod - 2, mod); }
int divi(int a, int b, int mod = 1e9 + 7) {
	return mul(a, inverse(b, mod), mod);
}
ll add(ll a, ll b, ll mod = 1e9 + 7) { return (((a % mod) + (b % mod)) + mod) % mod; }
ll sub(ll a, ll b, ll mod = 1e9 + 7) { return (((a % mod) - (b % mod)) + mod) % mod; }
ll mul(ll a, ll b, ll mod = 1e9 + 7) { return (((a % mod) * (b % mod)) + mod) % mod; }
ll bin(ll a, ll b, ll mod = 1e9 + 7) { ll ans = 1; while (b) { if (b & 1) ans = mul(ans, a, mod); a = mul(a, a, mod); b >>= 1; }return ans; }
ll inverse(ll a, ll mod = 1e9 + 7) { return bin(a, mod - 2, mod); }
ll divi(ll a, ll b, ll mod = 1e9 + 7) {
	return mul(a, inverse(b, mod), mod);
}
ll ex(int base, int power)
{
	if (power == 0)
		return 1;
	ll result = ex(base, power / 2);
	if (power % 2 == 1)
		return(((result * result) % mod) * base) % mod;
	else return (result * result) % mod;
}
int gcd(int a, int b) {
	if (b == 0)return a;
	else return gcd(b, a % b);
}
int lcm(int a, int b) {
	return a * b / gcd(a, b);
}
ll fac(int x) {
	ll factorial = 1;
	for (ll i = 1; i <= x; ++i) {
		factorial = mul(factorial, i);
	}
	return factorial % mod;
}
ll npr(int x, int c) {
	if (x == c) return fac(x);
	else return (fac(x) / (fac(x - c) * fac(c))) % mod;
}
void bton(string s) { stoll(s, nullptr, 2); }
inline int read() {
	int x = 0, f = 1, c = getchar();
	while (!isdigit(c)) { if (c == '-')f = -1; c = getchar(); }
	while (isdigit(c)) { x = (x << 1) + (x << 3) + (c ^ 48); c = getchar(); }
	return f == 1 ? x : -x;
}
bool isPrime(int n)
{
	if (n == 2 || n == 3)
		return true;

	if (n <= 1 || n % 2 == 0 || n % 3 == 0)
		return false;
	for (int i = 5; i * i <= n; i += 6) {
		if (n % i == 0 || n % (i + 2) == 0)
			return false;
	}

	return true;
}
int ceil_int(int first_number, int divider) {
	if (first_number % divider == 0) return first_number / divider;
	else return first_number / divider + 1;
}
ll ceil_ll(ll first_number, ll divider) {
	if (first_number % divider == 0) return first_number / divider;
	else return first_number / divider + 1LL;
}
/*
	t=1;
	first we can find a[0] and a[n-1] using the range 0,1e18+1 because a[0]>0 && a[n-1]<1e18+1
	then using that we can find a[1] by using a[0] because we know that a[1]>=a[0]+1 same for a[n-2]<=a[n-1]-1
	so can use MinMax once so that a[0] and a[n-1] is defined, then we can move into the middle so that we can recold {a[1],a[n-2]},{a[2],a[n-3]},....
	this solution will be n/2
	t=2;
	suggestion 1: binary search
	suggestion 2: skip the middle part when for example n=3 using math theory
	sugegstion 3: mx-mn
*/
#include "dna.h"
string aa, bb;
void init(std::string a, std::string b) {
	aa = a;
	bb = b;
}

int get_distance(int x, int y) {
	
	x--; y--;
	string abc;
	if (y - x == 0) return 0;
	else if (y - x == 1) {
		if (aa[x] == bb[x] && aa[y] == bb[y]) return 0;
		if (aa[x] == bb[y] && aa[y] == bb[x]) return 1;
		return -1;
	}
	else if (y - x == 2) {
		set<char> st1;
		FOR(i, x, y + 1) st1.insert(aa[i]);
		set<char> st2;
		FOR(i, x, y + 1) st2.insert(bb[i]);
		if (sz(st1) != sz(st2)) return -1;
		int ok = 1,cnt=0;
		FOR(i, x, y + 1) {
			if (aa[i] != bb[i]) ok = 0;
			else cnt++;
		}
		if (ok) return 0;
		else {
			return cnt - 1;
		}
	}
}

컴파일 시 표준 에러 (stderr) 메시지

dna.cpp: In function 'int get_distance(int, int)':
dna.cpp:137:9: warning: control reaches end of non-void function [-Wreturn-type]
  137 |  string abc;
      |         ^~~

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