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?
水題 會發現找規律後會發現答案就是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.
水題 循環所有行和列，求能使該行和該列形成十字需要添加的數的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.
先比對s和t字符串，然後看t中是否能使s變成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.
博弈論問題(以後一定要打表，自己分析很容易出錯)
會發現當k % 3 == 0 和k% 3 != 0 是不同的分開討論
- 當 k % 3 != 0 時會發現跳轉k，對最終的勝利數列沒有影響，都爲011的循環
- 當 k % 3 == 0 時
k = 3 時 01110111
k = 6 時 011011101101110110111
k = 9 時 01101101110110110111
所以時以k+1位數循環的n % (k + 1)) % 3 == 0 && n % (k + 1) != k時必爲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;
}