Educational Codeforces Round 157 (Rated for Div2)

A. Treasure Chest

思路：

当钥匙在宝箱左边时\(ans=x\);

当宝箱在钥匙左边时，若\(k>=y-x\)时，\(ans=y\),否则\(ans=2*y-x-k\)

code

int x, y, k;
void solved()
{
    cin >> x >> y >> k;
    if (x > y)
    {
        cout << x << endl;
    }
    else
    {
        cout << max(2 * y - x - k, y) << endl;
    }
}

B. Points and Minimum Distance

思路：

由图可总结出要最小值，即让\(x,y\)值相对集中，即对数组排序，前\(n\)做\(x\),后\(n\)做\(y\)

code:

int n, a[N];
void solved()
{
    cin >> n;
    for (int i = 1; i <= 2 * n; ++i)
    {
        cin >> a[i];
    }
    sort(a + 1, a + 1 + 2 * n);
    cout << a[n] + a[2 * n] - a[1] - a[n + 1] << endl;
    for (int i = 1; i <= n; ++i)
    {
        cout << a[i] << " " << a[i + n] << endl;
    }
}

C. Torn Lucky Ticket

思路：

分两种情况：一种是相同长度的连接，\(i,j\)可以互换，开数组m统计接相同值的字符串数量\(ans+=m[i]*m[i]\)

一种是不同长度的连接分别对字符串可以接的前后缀存在book数组中，\(ans+=book[i]*m[i]\)

会爆\(int\)

code:

int n;
void solved()
{
    cin >> n;
    int book[6][50] = {0};
    int m[6][50] = {0};
    for (int i = 0; i < n; ++i)
    {
        string s;
        cin >> s;
        if (s.size() == 5)
        {
            m[5][(s[0] - '0') + (s[1] - '0') + (s[2] - '0') + (s[3] - '0') + (s[4] - '0')]++;
            int tot = (s[0] - '0') + (s[1] - '0') + (s[2] - '0') + (s[3] - '0') - (s[4] - '0');
            book[3][tot]++;
            book[3][(s[4] - '0') + (s[1] - '0') + (s[2] - '0') + (s[3] - '0') - (s[0] - '0')]++;
            tot = (s[0] - '0') + (s[1] - '0') + (s[2] - '0') - (s[3] - '0') - (s[4] - '0');
            int op = (s[3] - '0') + (s[4] - '0') + (s[2] - '0') - (s[1] - '0') - (s[0] - '0');
            book[1][tot]++;
            book[1][op]++;
        }
        else if (s.size() == 4)
        {
            int tot = (s[0] - '0') + (s[1] - '0') + (s[2] - '0') + (s[3] - '0');
            m[4][tot]++;
            tot = (s[0] - '0') + (s[1] - '0') + (s[2] - '0') - (s[3] - '0');
            book[2][tot]++;
            int op = (s[3] - '0') + (s[1] - '0') + (s[2] - '0') - (s[0] - '0');
            book[2][op]++;
        }
        else if (s.size() == 3)
        {
            int tot = (s[0] - '0') + (s[1] - '0') + (s[2] - '0');
            m[3][tot]++;
            tot = (s[0] - '0') + (s[1] - '0') - (s[2] - '0');
            book[1][tot]++;
            int op = (s[1] - '0') + (s[2] - '0') - (s[0] - '0');
            book[1][op]++;
        }
        else if (s.size() == 2)
        {
            int tot = (s[0] - '0') + (s[1] - '0');
            m[2][tot]++;
        }
        else
        {
            int tot = (s[0] - '0');
            m[1][tot]++;
        }
    }
    int ans = 0;
    for (int i = 1; i <= 5; ++i)
    {
        for (int j = 0; j < 50; ++j)
        {
            if (book[i][j] != 0)
                ans += book[i][j] * m[i][j];
            ans += m[i][j] * m[i][j];
        }
    }
    cout << ans << endl;
}

D. XOR Construction

思路：

可根据\(b_{i}\oplus b_{i+1}=a_i\)可推出\(b_1\oplus b_i=a_1\oplus \dots \oplus a_i\)

则有\(b_1\oplus(a_1\oplus\dots\oplus a_i)=b_i\)

可以假设\(b_1\)的每一位为0，与\(a_i\)的前缀异或和异或，并与\(0\sim n-1\)异或并统计\(1\)的数目是否相等，若相等则假设正确，否则\(b_1\)的这一位为1

code:

int n, a[N];
void solved()
{
    cin >> n;
    for (int i = 1; i < n; ++i)
    {
        int tot;
        cin >> tot;
        a[i] = a[i - 1] ^ tot;
    }
    int x = 0;
    for (int i = 0; i < 30; ++i)
    {
        int p = 0, q = 0;
        for (int j = 0; j < n; ++j)
        {
            if (j & (1 << i))
                p++;
        }
        for (int j = 1; j < n; ++j)
        {
            if (a[j] & (1 << i))
                q++;
        }
        if (q != p)
            x += (1 << i);
    }
    cout << x << " ";
    for (int i = 1; i < n; ++i)
    {
        cout << (x ^ a[i]) << " ";
    }
    cout << endl;
}