题解:P11248 [GESP202409 七级] 矩阵移动

题解:洛谷 B4259 [GESP202503 二级] 等差矩阵 等知名刷题平台精心挑选而来,并结合各平台提供的算法标签和难度等级进行了系统分类。题目涵盖了从基础到进阶的多种算法和数据结构,旨在为不同阶段的编程学习者提供一条清晰、平稳的学习提升路径。列的矩阵,使得矩阵的每一行与每一列均是等差数列。小 A 发现,在矩阵的第。,得到的矩阵能满足要求。你能帮小 A 输出这个矩阵吗?个由空格分割的整数,表示小 A 需要构造的矩阵。本文分享的必刷题目是从。#入门# #语法基础# 阅读详情

题目传送门

题目大意

给出一个 n n n 行 m m m 列的只包含 0、1、? 的矩阵,你可以选择至多 x x x 个 ? 改成 1。

设得分为经过的 1 的数量,求从矩阵的 ( 1 , 1 ) (1,1) (1,1) 开始,每次只能向右或向下移动,走到 ( n , m ) (n,m) (n,m) 的最大得分为多少?

思路讲解

很容易想到动态规划。

设 d p i , j , k dp_{i,j,k} dpi,j,k​ 为从 ( 1 , 1 ) (1,1) (1,1) 走到 ( i , j ) (i,j) (i,j),修改路径上 k k k 个 ? 为 1 的最大得分, s i , j s_{i,j} si,j​ 为第 i i i 行 j j j 列的字符。

我们遍历 i , j , k i,j,k i,j,k,如果

GESP2024年9月认证C++七级真题解析 小杨开始在矩阵的左上角(1,1),小杨只能向下或者向右移动,最终到达右下角(n,m)时停止,在移动的过程中每经过一个字符1得分会增加一分(包括起点和终点),经过其它字符则分数不变。,先考虑当前有没有修改次数,如果没有当前阶段拿到和1和上一阶段比较求最大值,如果有修改次数,,需要看之前修改转移的值是多少,求最大值。【解析】在C++中,字符在计算机中是整数存储,小写字母b的ASCII码为98,对a进行自增操作时,其ASCII码增加到99,程序输出的是ASCII码值为99的字符,即小写字母c。 阅读详情

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