Codeforces Round #647 (Div. 2)

离谱前4题简单, 后两题难的离谱, E就过了150, F过了1个,

真就都卡在E做不下去

A

水题

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef double db;
 
const int N = 1e5 + 5;
 
int n, m, _, k;
 
int main()
{
    ios::sync_with_stdio(0); cin.tie(0);
    for (cin >> _; _; --_)
    {
        ll a, b; cin >> a >> b;
        if (a > b) swap(a, b);
        if (a == b) { cout << 0 << '\n'; continue; }
        if (b % a) {cout << -1 << '\n'; continue; } 
        ll c = b / a;
        int ans = 0;
        while (c % 8 == 0) ++ans, c /= 8;
        while (c % 4 == 0) ++ans, c /= 4;
        while (c % 2 == 0) ++ans, c /= 2;
        if (ans == 0 || c != 1) ans = -1;
        cout << ans << '\n';       
    }
    return 0;
}

B

暴力

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef double db;
 
const int N = 1e5 + 5;
 
int n, m, _, k;
int a[N], b[N];
 
int main()
{
    ios::sync_with_stdio(0); cin.tie(0);
    for (cin >> _; _; --_)
    {
        cin >> n;
        memset(b, 0, sizeof b);
        rep (i, 1, n) cin >> a[i], b[a[i]] = 1;
 
        int ans = 1, fl = 0;
        for (bool flag = 0; !flag && ans <= 1024; ++ans, fl = flag)
        {
            flag = 1;
            rep (i, 1, n)
                if (b[a[i] ^ ans] == 0) { flag = 0; break; }
        }
        if (fl) cout << ans - 1 << '\n';
        else cout << -1 << '\n';
    }
    return 0;
}

C

拆分二进制,先处理一下

然后从高位到低位,拆一算答案

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef double db;
 
const int N = 1e5 + 5;
 
int n, m, _, k;
ll s[65], f[65];
 
int main()
{
    ios::sync_with_stdio(0); cin.tie(0);
    s[0] = f[0] = 1;
    rep (i, 1, 60) s[i] = s[i - 1] << 1, f[i] = (f[i - 1] << 1) + i + 1;
    for (cin >> _; _; --_)
    {
        ll a, ans = 0; cin >> a;
        while (a)
        {
            if (a == 1) { ans += 1; break; }
            int idx = 0;
            while (s[idx] <= a) ++idx;
            //cout << idx << ' ' << s[idx - 1] << ' ' << f[idx - 2]  << '\n';
            ans += f[idx - 2] + idx; a -= s[idx - 1];
        }
        cout << ans << '\n';
    }
    return 0;
}

D

模拟完事

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef double db;
 
const int N = 5e5 + 5;
 
int n, m, _, k;
int h[N], to[N << 1], ne[N << 1], tot;
int a[N], ans[N], s[N], v[N];
 
void add(int u, int v)
{
    ne[++tot] = h[u]; h[u] = tot; to[tot] = v;
}
 
bool cmp(int x, int y)
{
    return a[x] < a[y];
}
 
int main()
{
    ios::sync_with_stdio(0); cin.tie(0);
    cin >> n >> m;
    rep (i, 1, m)
    {
        int u, v; cin >> u >> v;
        add(u, v); add(v, u);
    }
    rep(i, 1, n) cin >> a[i], s[i] = i;
 
    sort(s + 1, s + 1 + n, cmp);
 
    bool flag = 0;
    rep (i, 1, n)
    {
        //cout << i << ' ' << s[i] << ' ' << a[s[i]] << '\n';
        int x = s[i], c = 0;
        for (int j = h[x]; j; j = ne[j])
        {
            int y = to[j];
            if (a[y] == a[x]) { flag = 1; break; }
            if (a[y] < a[x] && v[a[y]] == 0) { v[a[y]] = 1, ++c; }
        }
        //if (i == 3) cout << flag << ' ' << c << '\n';
        if (flag || c < a[x] - 1) { flag = 1; break; }
        ans[i] = s[i];
        rep (j, 1, a[x] - 1) v[j] = 0;
    }
 
    if (flag) cout << -1;
    else  rep (i, 1, n) cout << ans[i] << ' ';
    return 0;
}