2020 计蒜之道 线上决赛 - C. 攀登山峰
Codeforces Round #289 (Div. 2) Tutorial
A. Maximum in Table
Simulation.
n = int(raw_input())
g = [[1 for i in xrange(n)] for j in xrange(n)]
for i in xrange(1, n):
for j in xrange(1, n):
g[i][j] = g[i - 1][j] + g[i][j - 1]
print g[n - 1][n - 1]
B …
more ...Codeforces Round #288 (Div. 2)
A. Pasha and Pixels
Brute force.
There are multiple ways to form a 2*2 square at one single step.
So at every step, we have to check the neighbours of pixel that is colored black.
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <vector>
using namespace std;
#define …Single Number Problem
Introduction
There are a lot of interview problem based on the 1D-array, which is the one of the easiest “data structure”.
But the problem about that simple data structure might not be that simple. Here is the summary of the problem about 1D-array.
Of course, most of them come from …
more ...Codeforces 447D DZY Loves Modification
Codeforces Round #242 (Div. 2) Tutorials and Solutions
A. Squats
Trun x => X or X => x to make the number of ‘x’ is equal to the number of ‘X’.
n = int(raw_input())
hamsters = [c for c in raw_input()]
sits = hamsters.count('x')
stands = hamsters.count('X')
if sits == stands:
print 0
print ''.join(hamsters)
else:
if sits > stands …Codeforces Round #240 (Div. 2) Tutorials and Solutions(incomplete and incorrect)
Overview
It has been months that I didn’t participate in the contest on CF, now I’m back. :)
This round of contest makes me confused that the problem B and C is a little bit too twisted, if you can’t catch the vital point, you will get a …
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