实验4 数组
TASK 1-1
#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"); for(i=0;i<N;++i) printf("%p:%d\n",&a[i],a[i]); printf("\n"); for(i=0;i<N;++i) printf("%p:%c\n",&b[i],b[i]); printf("\n"); printf("a=%p\n",a); printf("b=%p\n",b); return 0; }
①是,每个元素占用4内存字节单元;
②是,每个元素占用1内存字节单元;
③都是一样的。
TASK 1-2
#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; for(i=0;i<N;++i) for(j=0;j<M;++j) printf("%p:%d\n",&a[i][j],a[i][j]); printf("\n"); for(i=0;i<N;++i) for(j=0;j<M;++j) printf("%p:%c\n",&b[i][j],b[i][j]); return 0; }
①是,每个元素占用4个内存字节单元;
②是,每个元素占用1个内存字节单元。
TASK 2
#include<stdio.h> int days_of_year(int year,int month,int day); int main() { int year,month,day; int days; while(scanf("%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 a[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; int s=0,i; for(i=1;i<=month-1;++i) { s+=a[i]; } if(year%4==0&&year%100!=0) s+=1; return (s+day); }
TASK 3
#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("%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 i; double s=0.0; for(i=0;i<=n;i++) { s+=x[i]; } return (s/n); } void sort(int x[],int n) { int i,j,buf; for (i=0; i<n-1; ++i) { for (j=0; j<n-1-i; ++j) { if (x[j] < x[j+1]) { buf = x[j]; x[j] = x[j+1]; x[j+1] = buf; } } } }
TASK 4
#include<stdio.h> void dec2n(int x,int n); int main() { int x; printf("输入一个十进制整数:"); scanf("%d",&x); dec2n(x,2); dec2n(x,8); dec2n(x,16); return 0; } void dec2n(int x,int n) { int a[80]; int i=1,k,j=0; while(x!=0) { i=x%n; a[j++]=i; x=x/n; } for(k=j-1;k>=0;k--) { if(a[k]>9&&a[k]<16) printf("%c",a[k]-10+'A'); else printf("%d",a[k]); } printf("\n"); }
TASK 5
#include<stdio.h> int main() { int a[80][80]; int n,i,j,k,t; printf("Enter n:"); while(scanf("%d",&n)!=EOF) { for(i=0;i<n;i++) { for(j=i;j<n;j++) { a[i][j]=i+1; a[j][i]=i+1; } } for(k=0;k<n;k++) { for(t=0;t<n;t++) { printf("%d ",a[k][t]); } printf("\n"); } printf("\n"); printf("Enter n:"); } return 0; }
TASK 6
#include<stdio.h> #define N 80 int main() { char views1[N]="hey,c,i hate u."; char views2[N]="hey,c,i,love u."; int i; char buf[N]; printf("original views:\n"); printf("views1:%s\n",views1); printf("views2:%s\n",views2); printf("\n"); for(i=0;i<N;i++) { buf[i]=views1[i]; views1[i]=views2[i]; views2[i]=buf[i]; } printf("swapping...\n"); printf("views1:%s\n",views1); printf("views2:%s\n",views2); return 0; }
TASK 7
#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("输出初始名单:\n"); for(i=0;i<N;i++) printf("%s\n",name[i]); printf("\n排序中...\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 buf[N][M]; for(i=1;i<n;i++) { for(j=0;j<n-i;j++) { if(strcmp(str[j],str[j+1])>0) { strcpy(buf,str[j]); strcpy(str[j],str[j+1]); strcpy(str[j+1],buf); } } } }
实验总结
1.本次试验复习巩固了冒泡排序的算法;
学到了新的语句:strcmp(),我觉得这个语句非常非常实用;
实践任务五中灵光一闪找规律的思路惊喜到我自己了;
对于使用c语言转化进制的思路完善了。
2.仍旧是路漫漫其修远兮,加油。