Submission #1073462

#TimeUsernameProblemLanguageResultExecution timeMemory
1073462GrindMachineMutating DNA (IOI21_dna)C++17
100 / 100
50 ms7388 KiB
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a,b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a,b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(...) 42
#endif

/*



*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

#include "dna.h"

string s,t;

int f(char ch){
    if(ch == 'A') return 0;
    if(ch == 'C') return 1;
    if(ch == 'T') return 2; 
    assert(0);
    return -1;
}

int p[N][3], pc[N][3][3];

void init(std::string s, std::string t) {
    s = "$" + s, t = "$" + t;
    rep1(i,sz(s)-1){
        rep(j,3){
            p[i][j] = p[i-1][j]+(f(s[i]) == j)-(f(t[i]) == j);
            rep(k,3){
                if(j == k) conts;
                pc[i][j][k] = pc[i-1][j][k]+(f(s[i]) == j and f(t[i]) == k);
            }
        }
    }
}

int get_distance(int l, int r) {
    vector<int> cnt(3);
    int c[3][3];
    memset(c,0,sizeof c);

    rep(i,3){
        cnt[i] = p[r+1][i]-p[l][i];
        rep(j,3){
            c[i][j] = pc[r+1][i][j]-pc[l][i][j];
        }
    }

    rep(i,3) if(cnt[i]) return -1;

    int ans = 0;
    rep(i,3){
        for(int j = i+1; j < 3; ++j){
            int mn = min(c[i][j],c[j][i]);
            c[i][j] -= mn, c[j][i] -= mn;
            ans += mn;
        }
    }

    int mn = inf1, mx = -inf1;
    rep(i,3){
        rep(j,3){
            if(c[i][j]){
                amin(mn,c[i][j]);
                amax(mx,c[i][j]);
            }
        }
    }

    if(mn <= mx){
        assert(mn == mx);
        ans += 2*mn;
    }

    return ans;
}
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