제출 #1149550

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1149550		monkey133	Bliskost (COI23_bliskost)	C++20	100 / 100	883 ms	5584 KiB

Main

#include <bits/stdc++.h>
//#include "includeall.h"
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define endl '\n'
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define input(x) scanf("%lld", &x);
#define input2(x, y) scanf("%lld%lld", &x, &y);
#define input3(x, y, z) scanf("%lld%lld%lld", &x, &y, &z);
#define input4(x, y, z, a) scanf("%lld%lld%lld%lld", &x, &y, &z, &a);
#define print(x, y) printf("%lld%c", x, y);
#define show(x) cerr << #x << " is " << x << endl;
#define show2(x,y) cerr << #x << " is " << x << " " << #y << " is " << y << endl;
#define show3(x,y,z) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << endl;
#define all(x) x.begin(), x.end()
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
#define FOR(i, x, n) for (ll i =x; i<=n; ++i) 
#define RFOR(i, x, n) for (ll i =x; i>=n; --i) 
#pragma GCC optimize("O3","unroll-loops")
using namespace std;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
//using namespace __gnu_pbds;
//#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
//#define ordered_multiset tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>
typedef long long ll;
typedef long double ld;
typedef pair<ld, ll> pd;
typedef pair<string, ll> psl;
typedef pair<ll, ll> pi;
typedef pair<ll, pi> pii;
typedef pair<pi, pi> piii;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

ll n, q;
string ss, tt;
ll dif[1000005];
ll ans = 0;

int main()
{
	cin >> n >> q;
	cin >> ss >> tt;
	for (ll i=0; i<n; ++i) 
	{
		ll d = (ss[i] - tt[i] + 26)%26;
		ans = (ans + ((n-1-i)%2?-1:1)*d + 26)%26;
	}
	if (ans==0) cout << "da\n";
	else cout << "ne\n";
	while (q--)
	{
		ll idx;
		char ch;
		cin >> idx >> ch; idx--;
		ll d = (ss[idx] - tt[idx] + 26)%26;
		ans = (ans - ((n-1-idx)%2?-1:1)*d + 26)%26;
		ss[idx] = ch;
		d = (ss[idx] - tt[idx] + 26)%26;
		ans = (ans + ((n-1-idx)%2?-1:1)*d + 26)%26;
		if (ans==0) cout << "da\n";
		else cout << "ne\n";
	}

	return 0;
}

• Subtask 1: 7 / 7
• Subtask 2: 8 / 8
• Subtask 3: 13 / 13
• Subtask 4: 12 / 12
• Subtask 5: 17 / 17
• Subtask 6: 43 / 43