实验4
1.a是连续存放的,每个元素占用4个字节。
2.b是连续存放的,每个元素占用一个字节。
3.是一样的。
#include<stdio.h>
#define N 4
int main() {
int a[N] = { 2, 0, 2, 2 };
char b[N] = { '2', '0', '2', '2' };
int i; printf("sizeof(int) = %d\n", sizeof(int));
printf("sizeof(char) = %d\n", sizeof(char));
printf("\n"); // 输出数组a中每个元素的地址、值
for (i = 0; i < N; ++i)
printf("%p: %d\n", &a[i], a[i]);
printf("\n"); // 输出数组b中每个元素的地址、值
for (i = 0; i < N; ++i)
printf("%p: %c\n", &b[i], b[i]); printf("\n"); // 输出数组名a和b对应的值
printf("a = %p\n", a);
printf("b = %p\n", b);
return 0; }

1.是按行连续存放的,每个元素占4个字节
2.是按行连续存放的,每个元素占用一个字节
#include<stdio.h>
#define N 2
#define M 3
int main() {
int a[N][M] = { {1, 2, 3}, {4, 5, 6} };
char b[N][M] = { {'1', '2', '3'}, {'4', '5', '6'} };
int i, j; // 输出二维数组a中每个元素的地址和值
for (i = 0; i < N; ++i)
for (j = 0; j < M; ++j)
printf("%p: %d\n", &a[i][j], a[i][j]);
printf("\n"); // 输出二维数组a中每个元素的地址和值
for (i = 0; i < N; ++i)
for (j = 0; j < M; ++j)
printf("%p: %c\n", &b[i][j], b[i][j]);
return 0;
}

#include<stdio.h>
int days_of_year(int year,int month,int day);
int main()
{
int year,month,day;int days;
while(scanf_s("%d%d%d",&year,&month,&day)!=EOF)
{
days=days_of_year(year,month,day);
printf("%4d-%02d-%02d是这一年的第%d天.\n\n",year,month,day,days);
}
return 0;
}
int days_of_year(int year,int month,int day)
{
int i,days=0;
int mon[12]={0,31,28,31,30,31,30,31,31,30,31,30};
for(i=1;i<month;i++)
{
if((year%4==0)&&(year%100!=0)||year%400==0)
{
mon[2]++;
}
days+=mon[i];
}
days+=day;
return days;
}

#include<stdio.h>
#define N 5
void input(int x[],int n);
void output(int x[],int n);
double average(int x[],int n);
void sort(int x[],int n);
int main()
{
int scores[N];
double ave;
printf("录入%d个分数:\n",N);
input(scores,N);
printf("\n输出课程分数:\n");
output(scores,N);
printf("\n课程分数处理:计算均分、排序...\n");
ave=average(scores,N);
sort(scores,N);
printf("\n输出课程均分:%.2f\n",ave);
printf("\n输出课程分数(高—>低):\n");
output(scores,N);
return 0;
}
void input(int x[],int n)
{
int i;
for(i=0;i<n;++i)
scanf_s("%d",&x[i]);
}
void output(int x[],int n)
{
int i;
for(i=0;i<n;++i)
printf("%d ",x[i]);
printf("\n");
}
double average(int x[],int n)
{
int sum=0,i;double s;
for(i=0;i<n;++i)
{
sum+=x[i];
}
s=1.0*sum/n;
return s;
}
void sort(int x[],int n)
{
int i,j,k;
for(i=0;i<n;++i)
{
for(j=0;j<i;j++)
{
if(x[i]>x[j])
{
k=x[i];
x[i]=x[j];
x[j]=k;
}
}
}
}

#include<stdio.h>
void dec2n(int x, int n);
int main()
{
int x;
printf("输入一个十进制整数:");
scanf_s("%d", &x);
dec2n(x, 2);
printf("\n");
dec2n(x, 8);
printf("\n");
dec2n(x, 16);
printf("\n");
return 0;
}
void dec2n(int x, int n)
{
int a, b; char m[16] = { "0123456789ABCDEF" };
a = x / n;
b = x % n;
if (a == 0)
{
printf("%c", m[b]);
}
else
{
dec2n(a, n);
printf("%c", m[b]);
}
}

#include<stdio.h>
void ulmatrix(int n);
int main()
{
int n; int a[10][10];
printf("Enter n: ");
while (scanf_s("%d", &n) != EOF)
{
ulmatrix(n);
printf("\nEnter n: ");
}
return 0;
}
void ulmatrix(int n)
{
int i, j, a[10][10];
for (i = 0; i < n; i++)
{
for (j = 0; j < n; j++)
{
if (i <= j)
{
a[i][j] = i + 1;
}
else
{
a[i][j] = j + 1;
}
printf("%d ", a[i][j]);
}
printf("\n");
}
}
#include<stdio.h>
#define N 80
int main()
{
char views1[N]="hey,c,i hate u.";
char views2[N]="hey,c,i love u.";
char view[N];int i=0;
printf("original views:\n");
printf("views1: %s\n",views1);
printf("views2: %s\n",views2);
for(i=0;i<N;i++)
{
view[i]=views1[i];
views1[i]=views2[i];
views2[i]=view[i];
}
printf("\nswapping...\n\n");
printf("views1: %s\n",views1);
printf("views2: %s\n",views2);
return 0;
}

#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<string.h>
#define N 5
#define M 20
void bubble_sort(char str[][M], int n);
int main()
{
char name[][M] = { "Bob","Bill","Joseph","Taylor","George" };
int i;
printf("输入初始名单:\n");
for (i = 0; i < N; i++)
printf("%s\n", name[i]);
printf("\n排列中...\n");
bubble_sort(name, N);
printf("\n按字典序输出名单:\n");
for (i = 0; i < N; i++)
printf("%s\n", name[i]);
return 0;
}
void bubble_sort(char str[][M], int n)
{
int i, j; char mid[M];
for (i = 0; i < N; i++)
{
for (j = 0; j < i; j++)
{
if (strcmp(str[j], str[j + 1]) > 0)
{
strcpy(mid, str[j]);
strcpy(str[j], str[j + 1]);
strcpy(str[j + 1], mid);
}
}
}
}

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