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引言

        针对初学者，每日练习几个题，快速上手C语言。第十天。（连续更新中）

        采用在线OJ的形式

什么是在线OJ？

        在线判题系统（英语：Online Judge，缩写OJ）是一种在编程竞赛中用来测试参赛程序的在线系统，也可以用于平时的练习。

        详细内容可以看一下这篇博客：关于C/C++语言的初学者在哪刷题，怎么刷题-CSDN博客https://blog.csdn.net/2401_88433210/article/details/146056171?spm=1011.2415.3001.10575&sharefrom=mp_manage_link

 建议从第一天开始做，前面讲了的知识点后面只是会提一下。有不懂的写再评论区里 

        今天的题全是一些二维数组的应用

1.班级成绩输入输出 

        点进去直接做题：班级成绩输入输出_牛客题霸_牛客网 (nowcoder.com)https://www.nowcoder.com/practice/60d96b08e1cb42e38629d54e37eac008?tpId=107&&tqId=33382&rp=1&ru=/ta/beginner-programmers&qru=/ta/beginner-programmers/question-ranking参考代码：

#include<stdio.h>
int main()
{double arr[5][5];for(int i=0;i<5;++i){scanf("%lf %lf %lf %lf %lf",&arr[i][0],&arr[i][1],&arr[i][2],&arr[i][3],&arr[i][4]);}for(int i=0;i<5;++i){double sum=0;for(int j=0;j<5;++j){printf("%.1f ",arr[i][j]);sum+=arr[i][j];}printf("%.1f\n",sum);}return 0;
}

解析：按照题目要求做就行了

2.矩阵元素定位

        点进去直接多题：矩阵元素定位_牛客题霸_牛客网 (nowcoder.com)https://www.nowcoder.com/practice/b8e6a46992fe4e11b2822b20561b6d94?tpId=107&&tqId=33383&rp=1&ru=/ta/beginner-programmers&qru=/ta/beginner-programmers/question-ranking参考代码：

#include <stdio.h>int main() {int n, m;scanf("%d %d", &n, &m);int val[n + 1][m + 1];for (int i = 1; i <= n; i++) {for (int j = 1; j <= m; j++) {scanf("%d ", &val[i][j]);}}int x, y;scanf("%d %d", &x, &y);printf("%d", val[x][y]);return 0;
}

 解析：按照题目要求做就行了，简单题

3.序列重组矩阵

        点进去直接做题：序列重组矩阵_牛客题霸_牛客网 (nowcoder.com)https://www.nowcoder.com/practice/7548f8f5e44c42fa84fb2323d899a966?tpId=107&&tqId=33384&rp=1&ru=/ta/beginner-programmers&qru=/ta/beginner-programmers/question-ranking

 参考代码：

#include <stdio.h>int main()
{int n = 0;int m = 0;int arr[100] = { 0 };scanf("%d %d", &n, &m);for (int i = 0; i < n; i++){for (int j = 0; j < m; j++){scanf("%d", &arr[i]);printf("%d ", arr[i]);}printf("\n");}return 0;
}

解析：按照题目意思就行，简单的

4.最高身高

        点进去直接做题：最高身高_牛客题霸_牛客网 (nowcoder.com)https://www.nowcoder.com/practice/258fe0c567ac493f9b7bc9d3669d158d?tpId=107&&tqId=33385&rp=1&ru=/ta/beginner-programmers&qru=/ta/beginner-programmers/question-ranking参考代码：

#include <stdio.h>int main()
{int n = 0;int m = 0;int arr[10][10] = { 0 };scanf("%d %d", &n, &m);int max = 0;int x = 0;int y = 0;for (int i = 0; i < n; i++){for (int j = 0; j < m; j++){scanf("%d", &arr[i][j]);if (arr[i][j] > max){x = i + 1;y = j + 1;max = arr[i][j];}}}printf("%d %d\n", x, y);return 0;
}

解析：

        核心代码就if语句里面的东西，很简单。 

5.矩阵相等判定

        点进去直接做题：矩阵相等判定_牛客题霸_牛客网 (nowcoder.com)https://www.nowcoder.com/practice/2f058b1e89ae43f7ab650c0fd0b71fa8?tpId=107&&tqId=33386&rp=1&ru=/ta/beginner-programmers&qru=/ta/beginner-programmers/question-ranking参考代码：

#include <stdio.h>int main()
{int n = 0;int m = 0;int arr1[10][10] = { 0 };int arr2[10][10] = { 0 };//输入scanf("%d %d", &n, &m);int i = 0;for (i = 0; i < n; i++){for (int j = 0; j < m; j++){scanf("%d", &arr1[i][j]);}}for (i = 0; i < n; i++){for (int j = 0; j < m; j++){scanf("%d", &arr2[i][j]);//判断if (arr1[i][j] != arr2[i][j]){printf("No\n");return 0;}}}printf("Yes\n");return 0;
}

解析：题目让你干啥，你就干啥。

6.上三角矩阵判定

        点进去直接做题：
上三角矩阵判定_牛客题霸_牛客网 (nowcoder.com)https://www.nowcoder.com/practice/f5a29bacfc514e5a935723857e1245e4?tpId=107&&tqId=33387&rp=1&ru=/ta/beginner-programmers&qru=/ta/beginner-programmers/question-ranking参考代码：

//方法一：先将元素进行录入，后进行判别下三角的情况
int main()
{int n = 0;scanf("%d",&n);int i,j;int arr[10][10] = {0};for(i=0; i<n;i++){for(j=0;j<n;j++){scanf("%d",&arr[i][j]);}}for(i=0; i<n;i++){for(j=0;j<i;j++){if(arr[i][j] != 0){printf("NO\n");return 0;}}}printf("YES\n");return 0;
}
*///方法二：一边录入元素，一边进行判断
int main()
{int n = 0;scanf("%d",&n);int i,j;int arr[10][10] = {0};for(i=0; i<n;i++){for(j=0;j<n;j++){scanf("%d",&arr[i][j]);if(i>j){if(arr[i][j] != 0){//下三角有不是0的就不满足printf("NO");return 0;}}}}printf("YES\n");return 0;
}

解析：找规律，按照题意来做就行。

7.矩阵交换

        点进去直接做题：矩阵交换_牛客题霸_牛客网 (nowcoder.com)https://www.nowcoder.com/practice/ec44d4ff8c794b2f9205bdddbde96817?tpId=107&&tqId=33389&rp=1&ru=/ta/beginner-programmers&qru=/ta/beginner-programmers/question-ranking参考代码：

#include <stdio.h>int main()
{int n = 0;int m = 0;int arr[10][10] = { 0 };scanf("%d %d", &n, &m);for (int i = 0; i < n; i++){for (int j = 0; j < m; j++){scanf("%d", &arr[i][j]);}}int k = 0;scanf("%d", &k);char t = 0;int a = 0;int b = 0;//进行k次变化for (int i = 0; i < k; i++){scanf(" %c %d %d", &t, &a, &b);if (t == 'r'){//交换a和b行for (int j = 0; j < m; j++){int tmp = arr[a - 1][j];arr[a - 1][j] = arr[b - 1][j];arr[b - 1][j] = tmp;}}else if (t == 'c'){//交换a和b列for (int j = 0; j < n; j++){int tmp = arr[j][a - 1];arr[j][a - 1] = arr[j][b - 1];arr[j][b - 1] = tmp;}}}for (int i = 0; i < n; i++){for (int j = 0; j < m; j++){printf("%d ", arr[i][j]);}printf("\n");}return 0;
}

 解析：

1. 如果是%c读取字符，一定要清理前面输入时留下的空白字符

2. 注意二维数组的下标

3.按照题意模拟就行。

8.矩阵转置

        点进去直接做题：矩阵转置_牛客题霸_牛客网 (nowcoder.com)https://www.nowcoder.com/practice/351b3d03e410496ab5a407b7ca3fd841?tpId=107&&tqId=33388&rp=1&ru=/ta/beginner-programmers&qru=/ta/beginner-programmers/question-ranking参考代码：

#include <stdio.h>int main() {int m, n, b[20][20];scanf("%d %d", &n, &m);for (int i = 0; i < n; i++){for (int j = 0; j < m; j++){scanf("%d", &b[i][j]);}}for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {printf("%d ", b[j][i]);}printf("\n");}return 0;
}

解析：按照题目的意思做就行。

9.杨辉三角

        点进去直接做题：杨辉三角_牛客题霸_牛客网 (nowcoder.com)https://www.nowcoder.com/practice/e671c6a913d448318a49be87850adbcc?tpId=107&&tqId=33390&rp=1&ru=/ta/beginner-programmers&qru=/ta/beginner-programmers/question-ranking参考代码：

#include <stdio.h>int main()
{int n = 0;int arr[30][30] = { 0 };scanf("%d", &n);for (int i = 0; i < n; i++){for (int j = 0; j <= i; j++){if (0 == j || i == j){arr[i][j] = 1;}else{arr[i][j] = arr[i - 1][j - 1] + arr[i - 1][j];}printf("%5d", arr[i][j]);}printf("\n");}return 0;
}

解析：找规律题，模拟就行，注意题目要求。

10.井字棋

        点进去直接做题：井字棋_牛客题霸_牛客网 (nowcoder.com)https://www.nowcoder.com/practice/0375c1d206ef48a3894f72aa07b2fdde?tpId=107&&tqId=33391&rp=1&ru=/ta/beginner-programmers&qru=/ta/beginner-programmers/question-ranking参考代码：

#include <stdio.h>int main()
{char board[3][3] = { 0 };char flag = 0;for (int i = 0; i < 3; i++){for (int j = 0; j < 3; j++){scanf(" %c", &board[i][j]);}}//判断输赢for (int i = 0; i < 3; i++){//行if (board[i][0] == board[i][1] && board[i][1] == board[i][2] && board[i][2] != 'O'){flag = board[i][2];break;}//列if (board[0][i] == board[1][i] && board[1][i] == board[2][i] && board[2][i] != 'O'){flag = board[2][i];break;}}//对角线if (board[0][0] == board[1][1] && board[1][1] == board[2][2] && board[2][2] != 'O'){flag = board[2][2];}if (board[0][2] == board[1][1] && board[1][1] == board[2][0] && board[2][0] != 'O'){flag = board[2][0];}if ('K' == flag){printf("KiKi wins!\n");}else if ('B' == flag){printf("BoBo wins!\n");}else{printf("No winner!\n");}return 0;
}

 解析：

        把所有赢的情况列举出来然后排除就行。