Codeforces Round #647 (Div. 2)

题目传送门

A. Johnny and Ancient Computer

a经过一系列变换成为b需要的次数

简单的模拟

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, a, b) for (register int i = a; i <= b; i++)

ll a, b;
ll cnt;
void solve()
{
    cin >> a >> b;
    cnt = 0;
    while (a < b)
    {
        a *= 2;
        cnt++;
    }
    while (a > b && a % 2 == 0)
    {
        a /= 2;
        cnt++;
    }
    if (a != b)
        puts("-1");
    else if (cnt % 3 == 0)
        cout << cnt / 3 << endl;
    else
        cout << cnt / 3 + 1 << endl;
}
int main()
{
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
}

B. Johnny and His Hobbies

找到最小的k使 {s⊕k|s∈S}=S

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, a, b) for (register int i = a; i <= b; i++)

ll n, s[1100];
ll cnt[1100], tmp[1100];
void solve()
{
    cin >> n;
    memset(cnt, 0, sizeof(cnt));
    rep(i, 1, n)
    {
        cin >> s[i];
        cnt[s[i]]++;
    }
    rep(k, 1, 1023)
    {
        int flag = 1;
        rep(i, 0, 1023) tmp[i] = cnt[i];
        rep(i, 1, n)
        {
            tmp[s[i] ^ k]--;
            if (tmp[s[i] ^ k] < 0)
            {
                flag = 0;
                break;
            }
        }
        if (flag)
        {
            cout << k << endl;
            return;
        }
    }
    puts("-1");
}
int main()
{
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
}

C. Johnny and Another Rating Drop

很明显答案是单调的，一位一位的拆减就可以了

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, a, b) for (register int i = a; i <= b; i++)
ll n;
ll a[70], b[70];
ll ans;
void solve()
{
    for (ll i = 1, j = 1; j <= 1e18; i++, j *= 2)
    {
        a[i] = j;
        b[i] = b[i - 1] * 2 + 1;
    }
    cin >> n;
    ans = 0;
    for (int i = 60; i; i--)
        if (n >= a[i])
        {
            n -= a[i];
            ans += b[i];
        }
    cout << ans << endl;
}
int main()
{
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
}