Codeforces Round #731 (Div. 3) - 《写完题不来写题解？》

A Shortest Path with Obstacle">A Shortest Path with Obstacle
B Alphabetical Strings">B Alphabetical Strings
C Pair Programming">C Pair Programming
D Co-growing Sequence">D Co-growing Sequence
E Air Conditioners">E Air Conditioners
F Array Stabilization (GCD version)">F Array Stabilization (GCD version)

A Shortest Path with Obstacle

题目大意

给你一个位置A,一个位置B,还有一个不能经过的位置F，让你判断需要多少步能从A到B

主要思路

答案为A到B的曼哈顿距离，仅当ABF在一条直线上且F在AB中间时答案+2

AC代码

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <stack>
#include <string>
#include <cmath>
#include <unordered_map>
#include <queue>
#include <map>
using namespace std;
#define int long long
#define debug(a) cout << #a << " = " << a << endl;
#define x first
#define y second
typedef pair<int, int> P;
const int N = 200010;
int n, a[N];
signed main(void)
{
    int T;
    cin >> T;
    while(T--)
    {
        int x, y, x1, y1, x2, y2;
        cin >> x >> y >> x1 >> y1 >> x2 >> y2;
        if(x == x1 && x1 == x2 && (y < y2 && y2 < y1 || y > y2 && y2 > y1) || y == y1 && y1 == y2 && (x < x2 && x2 < x1 || x > x2 && x2 > x1))
        {
            cout << abs(x - x1) + abs(y - y1) + 2 << endl;
        }
        else
        {
            cout << abs(x - x1) + abs(y - y1) << endl;
        }
    }
    return 0;
}

B Alphabetical Strings

题目大意

从a到z添加字母，每次只能添加在字符串的最左边或者最右边，给定一个字符串，问该字符串是不是由此操作得到的

主要思路

从a开始，用两个指针来往左或往右判断即可

AC代码

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <stack>
#include <string>
#include <cmath>
#include <unordered_map>
#include <queue>
#include <map>
using namespace std;
#define int long long
#define debug(a) cout << #a << " = " << a << endl;
#define x first
#define y second
typedef pair<int, int> P;
const int N = 200010;
int n, a[N];
signed main(void)
{
    int T;
    cin >> T;
    while(T--)
    {
        string t;
        cin >> t;
        int l, r;
        bool flag1 = false;
        for(int i = 0; i < t.size(); i++)
        {
            if(t[i] == 'a') l = r = i, flag1 = true;
        }
        bool flag = true;
        char c = 'b';
        while(true)
        {
            if(l > 0 && t[l - 1] == c)
            {
                l--;
                c++;
            }
            else if(r < t.size() - 1 && t[r + 1] == c)
            {
                r++;
                c++;
            }
            else
            {
                if(l != 0 || r != t.size() - 1) flag = false;
                break;
            }
        }
        if(flag && flag1) puts("YES");
        else puts("NO");
    }
    return 0;
}

C Pair Programming

题目大意

刚开始给你一个大小K，然后给a，b两个数组，问你能否在不改变a，b数组本身顺序的情况下，合并a，b数组，且合并后的数组要满足，当数组元素为0时k+1，当不为0时该元素值>=k

主要思路

贪心+双指针，a，b两个数组分别一个指针，当有0时继续遍历，直到a,b都不为0时判断能否将两个元素放入合并后的数组

AC代码

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <stack>
#include <string>
#include <cmath>
#include <unordered_map>
#include <queue>
#include <map>
using namespace std;
#define int long long
#define debug(a) cout << #a << " = " << a << endl;
#define x first
#define y second
typedef pair<int, int> P;
const int N = 310;
int a[N], b[N];
int cnt, ans[N];
signed main(void)
{
    int T;
    cin >> T;
    while(T--)
    {
        cnt = 0;
        int k, n, m;
        cin >> k >> n >> m;
        for(int i = 0; i < n; i++) cin >> a[i];
        for(int i = 0; i < m; i++) cin >> b[i];
        int ta = 0, tb = 0;
        int t = k;
        bool flag = true;
        while(cnt < n + m && flag)
        {
            if(a[ta] == 0 && ta < n) ans[cnt++] = a[ta], ta++, t++;
            else if(b[tb] == 0 && tb < m) ans[cnt++] = b[tb], tb++, t++;
            else 
            {
                if(a[ta] <= t && ta < n) ans[cnt++] = a[ta], ta++;
                else if(b[tb] <= t && tb < m) ans[cnt++] = b[tb], tb++;
                else if(a[ta] > t && ta < n && b[tb] > t && tb < m ||
                (ta >= n && b[tb] > t && tb < m) ||
                (a[ta] > t && ta < n && tb >= m)) flag = false;
            }
        }
        if(flag)
        {
            for(int i = 0; i < n + m; i++)
            {
                cout << ans[i] << ' ';
            }
            cout << endl;
        }
        else cout << -1 << endl;
    }
    return 0;
}

D Co-growing Sequence

题目大意

给一个数组x，想让你求一个字典序最小数组y，使x的每个元素与y异或后的数组k，满足ki & ki+1 = ki

主要思路

当(x[i] & (x[i - 1] ^ y[i - 1])) == (x[i - 1] ^ y[i - 1])满足时y[i] = 0
否则，设t = x[i - 1] ^ y[i - 1]，(x ^ y) & t = t 说明t二进制表示中为1的位，(x ^ y)也一定为1，想求y最小本质上是找t中二进制位为1而x中二进制位为0的位，y中的这一位一定为1且仅这些位为1，那么y最小

AC代码

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <stack>
#include <string>
#include <cmath>
#include <unordered_map>
#include <queue>
#include <map>
using namespace std;
#define int long long
#define debug(a) cout << #a << " = " << a << endl;
#define x first
#define y second
typedef pair<int, int> P;
const int N = 200010;
int n, x[N], y[N];
int t[N];
signed main(void)
{
    int T;
    cin >> T;
    while(T--)
    {
        cin >> n;
        for(int i = 0; i < n; i++) cin >> x[i];
        y[0] = 0;
        for(int i = 1; i < n; i++)
        {
            if((x[i] & (x[i - 1] ^ y[i - 1])) == (x[i - 1] ^ y[i - 1])) y[i] = 0;
            else
            {
                int t = x[i - 1] ^ y[i - 1];
                y[i] = (x[i] ^ t) & t;
            }
        }
        for(int i = 0; i < n; i++) cout << y[i] << ' ';
        cout << endl;
    }
    return 0;
}

E Air Conditioners

题目大意

给一个n和k，位置为1~n，接下来给一个长度为k的数组表示空调的位置，再给一个长度为k的数组表示该位置空调的温度，距离空调位置为1那么温度+1，问每个位置的最低温度

主要思路

我们先从左往右扫描一遍空调对右边元素的影响，再从右往左扫描一遍空调对左边元素影响即可

AC代码

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <stack>
#include <string>
#include <cmath>
#include <unordered_map>
#include <queue>
#include <map>
using namespace std;
#define int long long
#define debug(a) cout << #a << " = " << a << endl;
#define x first
#define y second
typedef pair<int, int> P;
const int N = 300010;
int st[N], a[N];
int n, k;
signed main(void)
{
    int T;
    cin >> T;
    while(T--)
    {
        memset(a, 0x3f, sizeof(a));
        cin >> n >> k;
        for(int i = 0; i < k; i++) cin >> st[i];
        for(int i = 0; i < k; i++)
        {
            cin >> a[st[i]];
        }
        for(int i = 2; i <= n; i++) a[i] = min(a[i], a[i - 1] + 1);
        for(int i = n - 1; i >= 1; i--) a[i] = min(a[i], a[i + 1] + 1);
        for(int i = 1; i <= n; i++) cout << a[i] << ' ';
        puts("");
    }
    return 0;
}

F Array Stabilization (GCD version)

题目大意

给定一个长度为n的环，我们对数组每次操作都能使ai = gcd(a[i], a[i + 1])，问至少进行多少次操作使数组所有值相等

主要思路

将长度为n的环展开为2*n的数组,也就是将两个长度为n的数组拼起来，便于操作(环常用操作)。

gcd(gcd(a, b), gcd(b, c)) == gcd(gcd(a, b), c)

如果有两个相邻的数x[i], x[i+1]不等, 由于这个公式，那我们每次都往后取gcd直到两个数相同为止，记下每次操作的最大次数，最后取一个max即可。

AC代码

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <stack>
#include <string>
#include <cmath>
#include <unordered_map>
#include <queue>
#include <map>
using namespace std;
#define int long long
#define debug(a) cout << #a << " = " << a << endl;
#define x first
#define y second
typedef pair<int, int> P;
const int N = 200010;
int n, a[2 * N];
int gcd(int a, int b)
{
    return b ? gcd(b, a % b) : a;
}
signed main(void)
{
    int T;
    cin >> T;
    while(T--)
    {
        cin >> n;
        for(int i = 0; i < n; i++)
        {
            cin >> a[i];
            a[i + n] = a[i];
        }
        int ans = 0;
        for(int i = 0; i < n; i++)
        {
            int cnt = 0;
            int x = a[i];
            int y = a[i + 1];
            int t = i + 1;
            while(x != y)
            {
                x = gcd(x, a[t]);
                y = gcd(y, a[t + 1]);
                t++;
                cnt++;
            }
            ans = max(ans, cnt);
        }
        cout << ans << endl;
    }
    return 0;
}