Educational Codeforces Round 68

Educational Codeforces Round 68 (Rated for Div. 2)

A-Remove a Progression

You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?

水题 会发现找规律后会发现答案就是x*2

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
using namespace std;
#define ll long long
 
int main()
{
    int T;
    cin >> T;
    while (T--) {
        ll a, b;
        cin >> a >> b;
        cout << b * 2 << endl;
    }
    return 0;
}

B-Yet Another Crosses Problem

You are given a picture consisting of n rows and m columns. Rows are numbered from 1 to n from the top to the bottom, columns are numbered from 1 to m from the left to the right. Each cell is painted either black or white.
You think that this picture is not interesting enough. You consider a picture to be interesting if there is at least one cross in it. A cross is represented by a pair of numbers x and y, where 1≤x≤n and 1≤y≤m, such that all cells in row x and all cells in column y are painted black.
You have a brush and a can of black paint, so you can make this picture interesting. Each minute you may choose a white cell and paint it black.
What is the minimum number of minutes you have to spend so the resulting picture contains at least one cross?
You are also asked to answer multiple independent queries.

水题 循环所有行和列，求能使该行和该列形成十字需要添加的数的min

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
using namespace std;
#define ll long long
 
const int maxn = 50100;
int n, m;
int l[maxn], r[maxn];
string a[maxn];
 
int main()
{
    int T;
    scanf("%d", &T);
    while (T--) {
        memset(l, 0, sizeof(l));
        memset(r, 0, sizeof(r));
        scanf("%d%d", &n, &m);
        int maxl = 0;
        for (int i = 0; i < n; i++) {
            cin >> a[i];
            for (int j = 0; j < m; j++) {
                if (a[i][j] == '*') {
                    l[i]++;
                    r[j]++;
                }
            }
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                int t = l[i] + r[j];
                if (a[i][j] == '*') {
                    t--;
                }
                maxl = max(maxl, t);
            }
        }
        printf("%d\n", n + m - 1 - maxl);
    }
    return 0;
}

C-From S To T

You are given three strings s, t and p consisting of lowercase Latin letters. You may perform any number (possibly, zero) operations on these strings.
During each operation you choose any character from p, erase it from p and insert it into string s (you may insert this character anywhere you want: in the beginning of s, in the end or between any two consecutive characters).
Your goal is to perform several (maybe zero) operations so that s becomes equal to t. Please determine whether it is possible.
Note that you have to answer q independent queries.

先比对s和t字符串，然后看t中是否能使s变成t

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
using namespace std;
#define ll long long
 
const int maxn = 300;
char s[maxn], t[maxn], p[maxn];
int vis[maxn], vis2[maxn];
 
int main()
{
    int T;
    scanf("%d", &T);
    while (T--) {
        memset(vis, 0, sizeof(vis));
        memset(vis2, 0, sizeof(vis));
        scanf("%s", s);
        scanf("%s", t);
        scanf("%s", p);
        for (int i = 0; i < strlen(p); i++) {
            vis[p[i] - 'a']++;
        }
        int j = 0, flag = 0;
        for (int i = 0; i < strlen(s); i++) {
            if (flag)
                break;
            while (s[i] != t[j]) {
                j++;
                if (j > strlen(t)) {
                    flag = 1;
                    break;
                }
            }
            if (s[i] == t[j])
                vis2[j++] = 1;
        }
        for (int i = 0; i < strlen(t); i++) {
            if (flag)
                break;
            if (!vis2[i]) {
                if (vis[t[i] - 'a']) {
                    vis[t[i] - 'a']--;
                } else
                    flag = 1;
            }
        }
        if (flag)
            printf("NO\n");
        else
            printf("YES\n");
    }
    return 0;
}

D-1-2-K Game

Alice and Bob play a game. There is a paper strip which is divided into n + 1 cells numbered from left to right starting from 0. There is a chip placed in the n-th cell (the last one).
Players take turns, Alice is first. Each player during his or her turn has to move the chip 1, 2 or k cells to the left (so, if the chip is currently in the cell i, the player can move it into cell i - 1, i - 2 or i - k). The chip should not leave the borders of the paper strip: it is impossible, for example, to move it k cells to the left if the current cell has number i < k. The player who can’t make a move loses the game.
Who wins if both participants play optimally?
Alice and Bob would like to play several games, so you should determine the winner in each game.

博弈论问题(以后一定要打表，自己分析很容易出错)
会发现当k % 3 == 0 和k% 3 != 0 是不同的分开讨论

1. 当 k % 3 != 0 时会发现跳转k，对最终的胜利数列没有影响，都为011的循环
2. 当 k % 3 == 0 时
   k = 3 时 01110111
   k = 6 时 011011101101110110111
   k = 9 时 01101101110110110111
   所以时以k+1位数循环的n % (k + 1)) % 3 == 0 && n % (k + 1) != k时必为0

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
using namespace std;
#define ll long long
 
const int maxn = 300;
int n, k;
 
int main()
{
    int T;
    scanf("%d", &T);
    while (T--) {
        scanf("%d%d", &n, &k);
        if (k % 3 == 0) {
            if ((n % (k + 1)) % 3 == 0 && n % (k + 1) != k)
                printf("Bob\n");
            else
                printf("Alice\n");
            continue;
        }
        int t = n % 3;
        if (t == 0)
            printf("Bob\n");
        else
            printf("Alice\n");
    }
    return 0;
}