Codeforces Round #528 (Div. 2)题解

Codeforces Round #528 (Div. 2)题解

A. Right-Left Cipher

很明显这道题按题意逆序解码即可

Code：

# include <bits/stdc++.h>

int main()
{
    std::string s, t;
    std::cin >> s;
    int len = s.length();
    int cnt = 0;
    for(int i = 0; i < len; i++)
    {
        t = t + s[((len + 1) / 2 + cnt) - 1];
        if(i % 2 == 0)
            cnt = -cnt, ++cnt;
        else cnt = -cnt;
    }
    std::cout << t;
    return 0;
}

B. Div Times Mod

明显要使\(x\)最小，一定要使\(x \mod k\)最大

从\(n-1\)到\(1\)找能被\(n\)整除的最大数\(y\)

答案即为\((n/y)*k+y\)

Code：

# include <bits/stdc++.h>
# define ll long long
int main()
{
    ll n, k;
    ll ans = 0;
    scanf("%I64d%I64d", &n, &k);
    for(int i = 1; i < k; i++)
        if(n % i == 0)
            ans = i;
    printf("%I64d", (((n / ans) * k) + ans));
    return 0;
}

C. Connect Three

• 差点C没做出来退役

可以发现最优路径是曼哈顿距离上的两条路径

将\(3\)个点按\(x\)坐标排序，枚举(排序后的)\(A\)->\(B\)的两种(先上后左，先左后上)，\(B\)->\(C\)的两种，共四种

取最小值输出即可

Code：

#include <bits/stdc++.h>
#define mp(i, j) std::make_pair(i, j)
#define p std::pair<int, int>
p a[4];
std::map<p, int> m1, m2, m3, m4;
void add1()
{
    for (int i = a[1].first; i <= a[2].first; i++)
        m1[mp(i, a[1].second)] = 1;
    if (a[1].second <= a[2].second)
    {
        for (int i = a[1].second; i <= a[2].second; i++)
            m1[mp(a[2].first, i)] = 1;
    }
    else
    {
        for (int i = a[1].second; i >= a[2].second; i--)
            m1[mp(a[2].first, i)] = 1;
    }
    for (int i = a[2].first; i <= a[3].first; i++)
        m1[mp(i, a[2].second)] = 1;
    if (a[2].second <= a[3].second)
    {
        for (int i = a[2].second; i <= a[3].second; i++)
            m1[mp(a[3].first, i)] = 1;
    }
    else
    {
        for (int i = a[2].second; i >= a[3].second; i--)
            m1[mp(a[3].first, i)] = 1;
    }
}
void add2()
{
    for (int i = a[1].first; i <= a[2].first; i++)
        m2[mp(i, a[1].second)] = 1;
    if (a[1].second <= a[2].second)
    {
        for (int i = a[1].second; i <= a[2].second; i++)
            m2[mp(a[2].first, i)] = 1;
    }
    else
    {
        for (int i = a[1].second; i >= a[2].second; i--)
            m2[mp(a[2].first, i)] = 1;
    }

    if (a[2].second <= a[3].second)
    {
        for (int i = a[2].second; i <= a[3].second; i++)
            m2[mp(a[2].first, i)] = 1;
    }
    else
    {
        for (int i = a[2].second; i >= a[3].second; i--)
            m2[mp(a[2].first, i)] = 1;
    }
    for (int i = a[2].first; i <= a[3].first; i++)
        m2[mp(i, a[3].second)] = 1;
}
void add3()
{

    if (a[1].second <= a[2].second)
    {
        for (int i = a[1].second; i <= a[2].second; i++)
            m3[mp(a[1].first, i)] = 1;
    }
    else
    {
        for (int i = a[1].second; i >= a[2].second; i--)
            m3[mp(a[1].first, i)] = 1;
    }
    for (int i = a[1].first; i <= a[2].first; i++)
        m3[mp(i, a[2].second)] = 1;
    if (a[2].second <= a[3].second)
    {
        for (int i = a[2].second; i <= a[3].second; i++)
            m3[mp(a[2].first, i)] = 1;
    }
    else
    {
        for (int i = a[2].second; i >= a[3].second; i--)
            m3[mp(a[2].first, i)] = 1;
    }
    for (int i = a[2].first; i <= a[3].first; i++)
        m3[mp(i, a[3].second)] = 1;
}
void add4()
{
    if (a[1].second <= a[2].second)
    {
        for (int i = a[1].second; i <= a[2].second; i++)
            m4[mp(a[1].first, i)] = 1;
    }
    else
    {
        for (int i = a[1].second; i >= a[2].second; i--)
            m4[mp(a[1].first, i)] = 1;
    }
    for (int i = a[1].first; i <= a[2].first; i++)
        m4[mp(i, a[2].second)] = 1;
    for (int i = a[2].first; i <= a[3].first; i++)
        m4[mp(i, a[2].second)] = 1;
    if (a[2].second <= a[3].second)
    {
        for (int i = a[2].second; i <= a[3].second; i++)
            m4[mp(a[3].first, i)] = 1;
    }
    else
    {
        for (int i = a[2].second; i >= a[3].second; i--)
            m4[mp(a[3].first, i)] = 1;
    }
}
inline int print(std::map<p, int> m)
{
    for (std::map<p, int>::iterator it = m.begin(); it != m.end(); it++)
        printf("%d %d\n", it->first.first, it->first.second);
}
int main()
{
    int cnt = 0;
    for (int i = 1; i <= 3; i++)
        scanf("%d%d", &a[i].first, &a[i].second);
    std::sort(a + 1, a + 3 + 1);
    add1(), add2(), add3(), add4();
    cnt = std::min(std::min(m1.size(), m2.size()), std::min(m3.size(), m4.size()));
    printf("%d\n", cnt);
    if (m1.size() == cnt)
        return 0 * print(m1);
    if (m2.size() == cnt)
        return 0 * print(m2);
    if (m3.size() == cnt)
        return 0 * print(m3);
    if (m4.size() == cnt)
        return 0 * print(m4);
    return 0;
}