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35 changes: 35 additions & 0 deletions 2020-Autumn/Backend/task_02/LidongZou/2-1.cpp
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#include<stdio.h>
#include<stdlib.h>
void merge(int* a, int m, int* b, int n)
{
int i = m - 1, j = n - 1;
int k = m + n - 1;
while (i >= 0 && j >= 0)
{
if (a[i] >= b[j])
{
a[k--] = a[i];
i--;
}
else
{
a[k--] = b[j];
j--;
}
}
while (j >= 0)
{
a[k--] = b[j];
j--;
}
}
int main()
{
int a[] = { 1,2,3,0,0,0 };
int b[] = { 2,5,6 };
merge(a, 3, b, 3);
for (int i = 0; i < 6; i++)
printf("%d", a[i]);
system("pause");
return 0;
}
14 changes: 14 additions & 0 deletions 2020-Autumn/Backend/task_02/LidongZou/2-2.cpp
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#include<stdio.h>
void rotate(int** matrix, int n)
{
int i, j;
for (i = 1; i <= n; i++)
{
for (j = 1; j <= n; j++)
{
matrix[j][n - i + 1];

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这里是想做什么

printf_s("%d", matrix[j][n - i + 1]);
printf_s("\n");
}
}
}
8 changes: 8 additions & 0 deletions 2020-Autumn/Backend/task_02/LidongZou/2-3.cpp
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#include<stdio.h>
void traversal1(int* a, int n)
{
int echo[i];
a = echo;
for (n = 1; n <= i; n++)
printf_s("echo[%d]=%d\n", n, *(a + n))
}
15 changes: 15 additions & 0 deletions 2020-Autumn/Backend/task_02/LidongZou/2-4.cpp
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#include<stdio.h>
#include<stdlib.h>
int* createArray(int n)
{
int* a;
int* a = (int*)malloc(sizeof(int) * n);
scanf_s("%d", &n);
int i(1);
for (; i <= n; i++)
{
a[i] = i;
printf_s("%d", a[i]);
}
return 0;
}
18 changes: 18 additions & 0 deletions 2020-Autumn/Backend/task_02/LidongZou/2-5.cpp
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#include<stdio.h>
#include<stdlib.h>
int** createArray(int n, int m)
{
int** a;
int i, j;
a = (int**)malloc(sizeof(long int) * n);
for (i = 1; i <= m; i++)
{
for (j = 1; j <= n; j++)
{
a[i][j] = i * m - m + j;
printf_s("%d", a[i][j]);
}
printf_s("\n");
}
return 0;

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按题目要求请将创建好的数组返回

}
37 changes: 37 additions & 0 deletions 2020-Autumn/Backend/task_02/LidongZou/2-6.cpp
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#include<stdio.h>
int combination(int n, int m)
{
if (n == 1 && m == 1) return 1;
else if (n == 1 && m == 0) return 1;
else return combination(n - 1, m - 1) + combination(n - 1, m);
}
int main()
{
printf_s("Enter a number:");
int n, sum, t, i, a(0);
scanf_s("%d", &n);
if (n % 2 == 0)
{
sum = 0;
for (i = 0; i <= n / 2; i++)
{
t = combination(n / 2, i);
sum += t;
}
}

else
{
for (i = 0; i <= n / 2; i++)
{
sum = 0;
t = combination(n / 2, i);
t += (t + 1);
sum += t;
}
}
printf_s("%d", sum);
return 0;
}


7 changes: 7 additions & 0 deletions 2020-Autumn/Backend/task_02/LidongZou/README.md
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# 阴雨,考试,忙碌
##### 关于第二次作业的总结
这两周事情都很多,高数的期中考试,英语四级的口语考试,两套英语四级卷子,一套雅思试卷,1500字军事理论,高数做也做不完的习题。
上面的这些内容表面上是抱怨,实则我想说,忙碌可以证明自己的存在感,也顺便解释一下这次任务交的比较晚的原因。
对于数组,指针,我是属于零基础,在刚上手的时候,完全属于不知所措,看到题目有自己的想法之后,也有一些相关的语句,语法是自己所不知道的,也借助了百度的帮忙。也就只有第六题寻求了一下安慰。
那么话归正题,这一次作业给我的经验就是要在完成任务之前,要了解相关的知识,我在这里说的知识是更进一步的知识,更基础的知识,一些铺垫性质的知识,这样才能更好去完成任务。