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Codeforces Round #801 (Div. 2) and EPIC Institute of Technology Round C - Zero Path 在这道题目中,不可以真正地进行寻找必须另想办法. 考虑到只能向右面还有下面走,所以对于一个点,它只可能有已经走过的点转移过来,这让我想起来动态规划. 动态规划考虑最优解,这里所需要求的是0,不可以直接考虑是否与0相接近,需要考虑其他解法. 最大值以及最小值是唯一的,所以可以通过动态规划,最终求出目的地的最大数字以及最小数字. 证明…
Codeforces Round #801 (Div. 2) C(规律证明) 题目链接: 传送门QAQ 题意: 给定一个\(n * m\)的矩阵,矩阵的每个单元的值为1或-1,问从\((1,1)\)开始出发,每次只可以向下和向右走,问到终点\((n * m)\)时,是否可以总值为1. 分析: 题意很简单,本题的重点是在于,是否知道一个这样的结论: 首先定义 \(mn[i][j]\)为在\((i,j)\)处能拿到最少数量的1. \(mx[i][j]\)为在\((i,j)\)处能拿到最多数量的1.…
题集链接 A Subrectangle Guess 代码 #include <bits/stdc++.h> #define endl "\n" using namespace std; typedef long long ll; const int N = 1e6; void solve() { ll mis = -1e10; int a, b, n, m; cin >> n >> m; for (int i = 1; i <= n; i++)…
Codeforces Round #366 (Div. 2) A I hate that I love that I hate it水题 #I hate that I love that I hate it n = int(raw_input()) s = "" a = ["I hate that ","I love that ", "I hate it","I love it"] for i in ran…
Codeforces Round #354 (Div. 2) Problems     # Name     A Nicholas and Permutation standard input/output 1 s, 256 MB    x3384 B Pyramid of Glasses standard input/output 1 s, 256 MB    x1462 C Vasya and String standard input/output 1 s, 256 MB    x1393…
直达–>Codeforces Round #368 (Div. 2) A Brain’s Photos 给你一个NxM的矩阵,一个字母代表一种颜色,如果有”C”,”M”,”Y”三种中任意一种就输出”#Color”,如果只有”G”,”B”,”W”就输出”#Black&White”. #include <cstdio> #include <cstring> using namespace std; const int maxn = 200; const int INF =…
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Codeforces Round #262 (Div. 2) 1003 C. Present time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Little beaver is a beginner programmer, so informatics is his favorite subject. Soon his info…
Codeforces Round #262 (Div. 2) 1004 D. Little Victor and Set time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Little Victor adores the sets theory. Let us remind you that a set is a group of…