Educational Codeforces Round 47

A. Game Shopping

签.

#include <bits/stdc++.h>
using namespace std;

#define N 1010
int n, m, c[N];
queue <int> q;

int main()
{
    while (scanf("%d%d", &n, &m) != EOF)
    {
        while (!q.empty()) q.pop();
        for (int i = 1; i <= n; ++i) scanf("%d", c + i);
        for (int i = 1, x; i <= m; ++i) 
        {
            scanf("%d", &x);
            q.push(x);
        }
        int res = 0;
        for (int i = 1; i <= n && !q.empty(); ++i)
        {
            if (q.front() >= c[i])
            {
                q.pop();
                ++res;
            }
        }
        printf("%d
", res);
    }
    return 0;
}

B. Minimum Ternary String

签.

#include <bits/stdc++.h>
using namespace std;

#define N 100010
char s[N];
int n, cnt;

int main()
{
    while (scanf("%s", s + 1) != EOF)
    {
        n = strlen(s + 1);
        cnt = 0;
        for (int i = 1; i <= n; ++i) if (s[i] == '1')
            ++cnt;
        bool flag = false;
        for (int i = 1; i <= n; ++i)
        {
            if (flag && s[i] != '1')
               putchar(s[i]);
            else if (!flag)
            {
                if (s[i] == '0')
                    putchar(s[i]);
                else if (s[i] == '2')
                {
                    flag = true;
                    while (cnt--) putchar('1');
                    putchar(s[i]);
                }
            }    
        }
        if (cnt > 0) while (cnt--) putchar('1');
        puts("");
    }
    return 0;
}

C. Annoying Present

签.

#include <bits/stdc++.h>
using namespace std;

#define db long double
#define ll long long
#define N 100010
int n, m, x[N], d[N];

ll f(int x)
{
    return 1ll * n * (n + 1) / 2 + (1ll * x * x - 1ll * n * x - x);
}

int main()
{
    while (scanf("%d%d", &n, &m) != EOF)
    {
        ll Max = (ll)-1e18, Min = (ll)1e18;
        for (int i = 1; i <= n; ++i) 
        {
            ll tmp = f(i);
            Max = max(Max, tmp);
            Min = min(Min, tmp);
        }
        for (int i = 1; i <= m; ++i)
            scanf("%d%d", x + i, d + i);
        db res = 0;
        for (int i = 1; i <= m; ++i)
        {
            res += 1ll * n * x[i];
            if (d[i] < 0)
                res += 1ll * d[i] * Min;
            else 
                res += 1ll * d[i] * Max;
        }
        printf("%.10Lf
", res * 1.0 / n);
    }
    return 0;
}

D. Relatively Prime Graph

签。

$考虑质数的密度很大 并且phi(p) = p - 1 所以暴力找边即可$

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define N 100010
#define pii pair <int, int>
int n, m;
vector <pii> res;
int phi[N], prime[N];
bool check[N];
int tot;

void init()
{
    memset(check, false, sizeof check);
    phi[1] = 1;
    tot = 0;
    for (int i = 2; i < N; ++i)
    {
        if (!check[i])
        {
            prime[++tot] = i;
            phi[i] = i - 1;
        }
        for (int j = 1; j <= tot; ++j)
        {
            if (i * prime[j] >= N) break;
            check[i * prime[j]] = true;
            if (i % prime[j] == 0)
            {
                phi[i * prime[j]] = phi[i] * prime[j];
                break;
            }
            else
                phi[i * prime[j]] = phi[i] * (prime[j] - 1);
        }
    }
}

int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }

int main()
{
    init();
    while (scanf("%d%d", &n, &m) != EOF)
    {
        ll sum = 0;
        for (int i = 2; i <= n; ++i) sum += phi[i];
        if (m < n - 1 || m > sum) 
        {
            puts("Impossible");
            continue;
        }    
        puts("Possible");
        for (int i = 2; i <= n; ++i) printf("%d %d
", 1, i);
        m -= n - 1;
        vector <int> vec;
        for (int i = 3; i <= n; ++i) vec.push_back(i);
        sort(vec.begin(), vec.end(), [](int a, int b) { return phi[a] > phi[b]; });
        for (auto it : vec)
        {
            for (int i = 2; i < it && m > 0; ++i) if (gcd(it, i) == 1)
            {
                printf("%d %d
", it, i);
                --m;
            }
            if (m <= 0) break;
        }
    }
    return 0;
}

E. Intercity Travelling

签。

统计$a_i$被计算多少次

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define N 1000010
const ll p = (ll)998244353;
int n;
ll a[N], Bit[N];

int main()
{
    while (scanf("%d", &n) != EOF)
    {
        for (int i = 1; i <= n; ++i) scanf("%lld", a + i);
        Bit[0] = 1;
        for (int i = 1; i <= n; ++i) Bit[i] = (Bit[i - 1] * 2) % p;
        if (n == 1)
        {
            printf("%lld
", a[1]); 
            continue;
        }
        ll res = 0;
        for (int i = 1; i <= n; ++i)
        {
            res = (res + Bit[n - i] * a[i] % p) % p;
            if (n - i - 1 >= 0)
                res = (res + 1ll * (n - i) * Bit[n - i - 1] % p * a[i] % p) % p;
        }
        printf("%lld
", res);
    }
    return 0;
}