From 753ef224e3103ea6208ee6de91d597d1e9466b56 Mon Sep 17 00:00:00 2001 From: root Date: Sun, 24 Nov 2024 15:33:02 +0800 Subject: [PATCH] =?UTF-8?q?=E4=BA=8C=E6=9C=9F=E6=A0=A1=E8=B5=9B?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- output/一桌多少人 | Bin 0 -> 43544 bytes output/口矩阵 | Bin 0 -> 44792 bytes output/数组偶数最小值 | Bin 0 -> 43768 bytes output/是否有字符串 | Bin 0 -> 45168 bytes 一桌多少人.c | 40 +++++++++ 口矩阵.c | 106 ++++++++++++++++++++++ 寻宝.c | 205 ++++++++++++++++++++++++++++++++++++++++++ 数组偶数最小值.c | 46 ++++++++++ 是否有字符串.c | 54 +++++++++++ 9 files changed, 451 insertions(+) create mode 100755 output/一桌多少人 create mode 100755 output/口矩阵 create mode 100755 output/数组偶数最小值 create mode 100755 output/是否有字符串 create mode 100644 一桌多少人.c create mode 100644 口矩阵.c create mode 100644 寻宝.c create mode 100644 数组偶数最小值.c create mode 100644 是否有字符串.c diff --git a/output/一桌多少人 b/output/一桌多少人 new file mode 100755 index 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zB7d{llsCWAY(i+Ga+&^{d?tJcQu!Ou=H2`bvO#Sz_6%)q05XPsLUF^U4a8y;?oGeEB$2 z6s)+$Gb%K9Q{IH<9HRUeu2v;ZG!t7_V9J^Bd|lquZ^D1MMm_z62CVNg_kKM!YV^pP@e^Kgr+9P$UI+eL@0jeOcw@?bAH3RDR#ztMdEG3Y5KC{};+N Bdy@bF literal 0 HcmV?d00001 diff --git a/一桌多少人.c b/一桌多少人.c new file mode 100644 index 0000000..f989d8b --- /dev/null +++ b/一桌多少人.c @@ -0,0 +1,40 @@ +// 【问题描述】 +// 小蓝准备请自己的朋友吃饭。小蓝朋友很多,最终吃饭的人总数达 2024 人(包括他自己)。 + +// 请问如果每桌最多坐 n 人,最少要多少桌才能保证每个人都能吃饭。 + +// 【输入格式】 +// 输入一行包含一个整数 n 。 + +// 【输出格式】 +// 输出一行包含一个整数,表示最少的桌数。 + +// 【样例输入】 +// 10 + +// 【样例输出】 +// 203 + +// 【样例输入】 +// 8 + +// 【样例输出】 +// 253 + +// 【评测用例规模与约定】 +// 对于所有评测用例,1 <= n <= 2024。 + + + +#include + +int main() { + int n; + scanf("%d", &n); + int result = 2024 / n; + if (2024 % n != 0) { + result++; + } + printf("%d\n", result); + return 0; +} \ No newline at end of file diff --git a/口矩阵.c b/口矩阵.c new file mode 100644 index 0000000..6349a62 --- /dev/null +++ b/口矩阵.c @@ -0,0 +1,106 @@ +// 问题描述】 +// 小蓝有一个 n 行 m 列的矩阵 a[i][j] ,他想在矩阵中找出一个“口”字形状的区域,使得区域上的值的和最大。 + +// 具体讲,一个“口”字形状的区域可以由两个坐标 (x1, y1) 和 (x2, y2) 确定,满足: + +// 1 <= x1 < x2 <= n ; +// 1 <= y1 < y2 <= m ; +// x2 - x1 = y2 - y1 。 +// 对应的区域由满足以下条件之一的点 (x, y) 构成: + +// x1 <= x <= x2,且 y = y1 ,对应“口”的左边一竖; +// y1 <= y <= y2,且 x = x1 ,对应“口”的上面一横; +// x1 <= x <= x2,且 y = y2 ,对应“口”的右边一竖; +// y1 <= y <= y2,且 x = x2 ,对应“口”的下面一横。 +// 请注意有些点满足以上条件的多个,例如左上角的点 (x1, y1) ,在计算时算为一个点。 + +// 区域上的值是指对应区域的所有点的值,即“口”字的框上的值,不含框内和框外的值。 + +// 【输入格式】 +// 输入的第一行包含两个整数 n, m ,分别表示行数和列数。 + +// 接下来 n 行,每行包含 m 个整数,相邻数之间使用一个空格分隔,依次表示矩阵的每行每列的值,本部分的第 i 行第 j 列表示 a[i][j] 。 + +// 【输出格式】 +// 输出一行包含一个整数,表示最大的和。 + +// 【样例输入】 +// 5 6 +// 1 -1 2 -2 3 -3 +// -1 2 -2 3 -3 4 +// 2 -2 3 -3 4 -4 +// -2 3 -3 4 -4 5 +// 3 -3 4 -4 5 -5 +// 【样例输出】 +// 4 +// 【样例说明】 +// 取 (x1, y1) = (1, 1) , (x2, y2) = (5, 5) 可得到最大值。 + +// 【评测用例规模与约定】 +// 对于 30% 的评测用例,1 <= n, m <= 30 ,-1000 <= a[i][j] <= 1000 。 + +// 对于 60% 的评测用例,1 <= n, m <= 100 ,-1000 <= a[i][j] <= 1000 。 + +// 对于所有评测用例,1 <= n, m <= 300 ,-1000 <= a[i][j] <= 1000 。 + +#include + +#define MAX 301 + +int a[MAX][MAX]; +int row_sum[MAX][MAX]; +int col_sum[MAX][MAX]; + +int main() { + int n, m; + scanf("%d %d", &n, &m); + + for (int i = 1; i <= n; i++) { + for (int j = 1; j <= m; j++) { + scanf("%d", &a[i][j]); + } + } + + for (int i = 1; i <= n; i++) { + row_sum[i][0] = 0; + for (int j = 1; j <= m; j++) { + row_sum[i][j] = row_sum[i][j - 1] + a[i][j]; + } + } + + for (int j = 1; j <= m; j++) { + col_sum[j][0] = 0; + for (int i = 1; i <= n; i++) { + col_sum[j][i] = col_sum[j][i - 1] + a[i][j]; + } + } + + int max_sum = -1000000000; + + int k_max = (n - 1 < m - 1) ? (n - 1) : (m - 1); + + for (int k = 1; k <= k_max; k++) { + for (int x1 = 1; x1 <= n - k; x1++) { + int x2 = x1 + k; + for (int y1 = 1; y1 <= m - k; y1++) { + int y2 = y1 + k; + + int sum_left = col_sum[y1][x2] - col_sum[y1][x1 - 1]; + int sum_top = row_sum[x1][y2] - row_sum[x1][y1 - 1]; + int sum_right = col_sum[y2][x2] - col_sum[y2][x1 - 1]; + int sum_bottom = row_sum[x2][y2] - row_sum[x2][y1 - 1]; + + int sum_corners = a[x1][y1] + a[x1][y2] + a[x2][y1] + a[x2][y2]; + + int total = sum_left + sum_top + sum_right + sum_bottom - sum_corners; + + if (total > max_sum) { + max_sum = total; + } + } + } + } + + printf("%d\n", max_sum); + return 0; +} \ No newline at end of file diff --git a/寻宝.c b/寻宝.c new file mode 100644 index 0000000..0a1075c --- /dev/null +++ b/寻宝.c @@ -0,0 +1,205 @@ +// 【问题描述】 +// 小蓝正在玩一个寻宝游戏。寻宝游戏在一个方格图上进行。方格图中的每一个格子都有一个坐标 (r, c),其中越往北 r 越小,越往南 r 越大,越往东 c 越大,越往西 c 越小。南北相邻方格的 c 坐标相同,r 坐标差一。东西相邻方格的 r 坐标相同, c 坐标差一。 + +// 游戏开始时,小蓝站在 (0, 0) 处,面向北边。游戏区域无限大,且没有障碍。每一步,小蓝控制自己的角色走一步,他可以有如下三种选择: + +// 向前走:朝现在的方向前进到相邻的方格中,并保持当前的方向。 +// 向左走:向左转90度,并前进到相邻的方格中(即进入到原来左边的方格),面向的方向变为了原来的左边。 +// 向右走:向右转90度,并前进到相邻的方格中(即进入到原来右边的方格),面向的方向变为了原来的右边。 +// 小蓝玩了一会儿,一共走了 n 步,他记录了自己的每一个动作。但是他没有找到宝藏。他怀疑前面的某一步出现了失误。他想知道,如果他改变之前的某一步,能到的位置有哪些。由于这个太复杂,他想知道最终到的位置(第 n 步后到的位置)有多少种。 + +// 【输入格式】 +// 输入的第一行包含一个整数 n ,表示小蓝走了 n 步。 + +// 第二行包含一个长度为 n 的由大写字母组成的字符串,依次表示小蓝走的每一步。字母 F 、 L 、 R 分别表示对应的步是向前走、向左走、向右走。 + +// 【输出格式】 +// 输出一行,包含一个整数,表示如果改变某一步,可以到的位置的种类数。 + +// 【样例输入】 +// 3 +// FLR +// 【样例输出】 +// 6 +// 【样例说明】 +// 如果不改变,三步依次走到:(-1, 0), (-1, -1), (-2, -1) ,最终位置为 (-2, -1) 。 + +// 如果第一步改成 L,三步依次走到:(0, -1), (1, -1), (1, -2) ,最终位置为 (1, -2) 。 + +// 如果第一步改成 R,三步依次走到:(0, 1), (-1, 1), (-1, 2) ,最终位置为 (-1, 2) 。 + +// 如果第二步改成 F,三步依次走到:(-1, 0), (-2, 0), (-2, 1) ,最终位置为 (-2, 1) 。 + +// 如果第二步改成 R,三步依次走到:(-1, 0), (-1, 1), (0, 1) ,最终位置为 (0, 1) 。 + +// 如果第三步改成 F,三步依次走到:(-1, 0), (-1, -1), (-1, -2) ,最终位置为 (-1, -2) 。 + +// 如果第三步改成 L,三步依次走到:(-1, 0), (-1, -1), (0, -1) ,最终位置为 (0, -1) 。 + +// 共有 6 种不同的最终位置。 + +// 【样例输入】 +// 4 +// RRRR +// 【样例输出】 +// 6 +// 【样例说明】 +// 有 8 种改变方法: + +// 改为 FRRR ,最终位置为 (0, 0); + +// 改为 LRRR ,最终位置为 (0, 0); + +// 改为 RFRR ,最终位置为 (1, 1); + +// 改为 RLRR ,最终位置为 (0, 2); + +// 改为 RRFR ,最终位置为 (2, 0); + +// 改为 RRLR ,最终位置为 (2, 2); + +// 改为 RRRF ,最终位置为 (1, -1); + +// 改为 RRRL ,最终位置为 (2, 0)。 + +// 不同的最终位置共有 6 种。 + +// 【评测用例规模与约定】 +// 对于 30% 的评测用例,1 <= n <= 20。 + +// 对于 60% 的评测用例,1 <= n <= 1000。 + +// 对于所有评测用例,1 <= n <= 100000 + +#include +#include + +#define MAXN 100000 + +int n; +char actions[MAXN + 1]; +int original_dir[MAXN + 1]; +int original_pos_x[MAXN + 1]; +int original_pos_y[MAXN + 1]; + +int dx[4] = {-1, 0, 1, 0}; +int dy[4] = {0, 1, 0, -1}; + +void simulate(int step, int new_action, int *final_x, int *final_y) { + int dir = original_dir[step]; + int x = original_pos_x[step]; + int y = original_pos_y[step]; + + if (new_action == 'F') { + // 前进,方向不变 + } else if (new_action == 'L') { + // 向左转,方向减一 + dir = (dir - 1 + 4) % 4; + } else if (new_action == 'R') { + // 向右转,方向加一 + dir = (dir + 1) % 4; + } + + // 执行新的动作 + x += dx[dir]; + y += dy[dir]; + + // 继续模拟后续步骤 + for (int i = step + 1; i < n; i++) { + char action = actions[i]; + if (action == 'F') { + // 前进,方向不变 + } else if (action == 'L') { + // 向左转,方向减一 + dir = (dir - 1 + 4) % 4; + } else if (action == 'R') { + // 向右转,方向加一 + dir = (dir + 1) % 4; + } + x += dx[dir]; + y += dy[dir]; + } + + *final_x = x; + *final_y = y; +} + +int main() { + scanf("%d", &n); + scanf("%s", actions); + + // 初始化原始方向和位置 + original_dir[0] = 0; + original_pos_x[0] = 0; + original_pos_y[0] = 0; + + // 预计算原始方向和位置序列 + for (int i = 0; i < n; i++) { + char action = actions[i]; + int dir = original_dir[i]; + int x = original_pos_x[i]; + int y = original_pos_y[i]; + + if (action == 'F') { + // 前进,方向不变 + } else if (action == 'L') { + // 向左转,方向减一 + dir = (dir - 1 + 4) % 4; + } else if (action == 'R') { + // 向右转,方向加一 + dir = (dir + 1) % 4; + } + + // 更新位置 + x += dx[dir]; + y += dy[dir]; + + original_dir[i + 1] = dir; + original_pos_x[i + 1] = x; + original_pos_y[i + 1] = y; + } + + // 使用哈希表存储最终位置 + int hash[200001][2]; + int hash_size = 0; + + for (int i = 0; i < n; i++) { + char original_action = actions[i]; + // 可能的动作变化 + char possible_actions[2]; + if (original_action == 'F') { + possible_actions[0] = 'L'; + possible_actions[1] = 'R'; + } else if (original_action == 'L') { + possible_actions[0] = 'F'; + possible_actions[1] = 'R'; + } else if (original_action == 'R') { + possible_actions[0] = 'F'; + possible_actions[1] = 'L'; + } + + for (int j = 0; j < 2; j++) { + char new_action = possible_actions[j]; + int final_x, final_y; + simulate(i, new_action, &final_x, &final_y); + + // 检查是否已存在该位置 + int found = 0; + for (int k = 0; k < hash_size; k++) { + if (hash[k][0] == final_x && hash[k][1] == final_y) { + found = 1; + break; + } + } + if (!found) { + hash[hash_size][0] = final_x; + hash[hash_size][1] = final_y; + hash_size++; + } + } + } + + printf("%d\n", hash_size); + + return 0; +} \ No newline at end of file diff --git a/数组偶数最小值.c b/数组偶数最小值.c new file mode 100644 index 0000000..6f45e82 --- /dev/null +++ b/数组偶数最小值.c @@ -0,0 +1,46 @@ +// 【问题描述】 +// 小蓝有一个数组 a[1], a[2], ..., a[n] ,请求出数组中值最小的偶数,输出这个值。 + +// 【输入格式】 +// 输入的第一行包含一个整数 n 。 + +// 第二行包含 n 个整数,相邻数之间使用一个空格分隔,依次表示 a[1], a[2], ..., a[n] 。 + +// 【输出格式】 +// 输出一行,包含一个整数,表示答案。数据保证数组中至少有一个偶数。 + +// 【样例输入】 +// 9 +// 9 9 8 2 4 4 3 5 3 +// 【样例输出】 +// 2 +// 【样例输入】 +// 5 +// 4321 2143 1324 1243 4312 +// 【样例输出】 +// 1324 +// 【评测用例规模与约定】 +// 对于 30% 的评测用例,1 <= n <= 100,0 <= a[i] <= 1000。 + +// 对于 60% 的评测用例,1 <= n <= 1000,0 <= a[i] <= 1000。 + +// 对于所有评测用例,1 <= n <= 10000,0 <= a[i] <= 1000000。 + +#include + +int main() { + int n; + scanf("%d", &n); + int a[n]; + for(int i = 0; i < n; i++) { + scanf("%d", &a[i]); + } + int min_even = 1000001; + for(int i = 0; i < n; i++) { + if(a[i] % 2 == 0 && a[i] < min_even) { + min_even = a[i]; + } + } + printf("%d\n", min_even); + return 0; +} \ No newline at end of file diff --git a/是否有字符串.c b/是否有字符串.c new file mode 100644 index 0000000..25c9484 --- /dev/null +++ b/是否有字符串.c @@ -0,0 +1,54 @@ +// 【问题描述】 +// 一个字符串包含LANQIAO是指在字符串中能取出几个字符,将他们按照在原串中的位置顺序摆成一排后字符串为 LANQIAO 。即字符串包含 LANQIAO 是指 LANQIAO 是这个串的子序列。 + +// 例如:LLLLLANHAHAHAQLANIIIIALANO 中包含 LANQIAO 。 + +// 又如:OAIQNAL 中不包含 LANQIAO 。 + +// 给点一个字符串,判断字符串中是否包含 LANQIAO 。 + +// 【输入格式】 +// 输入一行包含一个字符串。 + +// 【输出格式】 +// 如果包含 LANQIAO ,输出一个英文单词 YES ,否则输出一个英文单词 NO 。 + +// 【样例输入】 +// LLLLLANHAHAHAQLANIIIIALANO +// 【样例输出】 +// YES +// 【样例输入】 +// OAIQNAL +// 【样例输出】 +// NO +// 【评测用例规模与约定】 +// 对于所有评测用例,输入的字符串非空串,由大写字母组成,长度不超过 1000 。 + +#include +#include + +int main() { + char s[1001]; + fgets(s, 1001, stdin); + // 去掉换行符 + size_t len = strlen(s); + if (len > 0 && s[len-1] == '\n') { + s[len-1] = '\0'; + } + char t[] = "LANQIAO"; + int j = 0; + for (int i = 0; s[i] != '\0'; i++) { + if (s[i] == t[j]) { + j++; + if (j == 7) { + break; + } + } + } + if (j == 7) { + printf("YES\n"); + } else { + printf("NO\n"); + } + return 0; +} \ No newline at end of file