实验4
实验任务1
#include <stdio.h> const int N = 4; int main(){ int a[N]={2,0,2,1}; char b[N]={'2','0','1','1'}; 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("%x:%d\n",&a[i],a[i]); printf("\n"); for(i=0;i<N;++i) printf("%x:%c\n",&b[i],b[i]); return 0; }
数组a中,内存是连续存放的,每个元素占4个字节。
数组b中,内存是连续存放的,每个元素占4个字节。
#include <stdio.h> int main(){ int a[2][3] = {{1, 2, 3}, {4, 5, 6}}; char b[2][3] = {{'1', '2', '3'}, {'4', '5', '6'}}; int i, j; for (i = 0; i < 2; ++i) for (j = 0; j < 3; ++j) printf("%x: %d\n", &a[i][j], a[i][j]); printf("\n"); for (i = 0; i < 2; ++i) for (j = 0; j < 3; ++j) printf("%x: %c\n", &b[i][j], b[i][j]); }
a中不完全连续,占4字节
b中连续,占1字节
实验任务2
#include <stdio.h> #define N 1000 int fun(int n,int m,int bb[N]) { int i, j, k = 0, flag; for (j = n; j <= m; j++){ flag=1; for (i = 2; i < j; i++) if (j%i==0) { flag = 0; break; } if (flag!=0) bb[k++] = j; } return k; } int main() { int n = 0, m = 0, i, k, bb[N]; scanf("%d", &n); scanf("%d", &m); for (i = 0; i < m - n; i++) bb[i] = 0; k = fun(n,m,bb); for (i = 0; i < k; i++) printf("%4d", bb[i]); return 0; }
实验任务3
#include <stdio.h> const int N = 5; int find_max(int x[], int n); void input(int x[], int n); void output(int x[], int n); int main(){ int a[N]; int max; input(a, N); output(a, N); max = find_max(a, N); printf("max = %d\n", max); return 0; 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"); } int find_max(int x[], int n) { int max=0; for(int i=0;i<n;i++) { if(x[i]>max) max=x[i]; } return max; }
实验任务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 k; int m=0; k=x; while(k){ k=k/n; m++; } int a[m]; for(int i=0;i<m;i++){ a[i]=x%n; x=x/n; } for(int i=m-1;i>=0;i--) switch(a[i]){ case 10: printf("A");break; case 11: printf("B");break; case 12: printf("C");break; case 13: printf("D");break; case 14: printf("E");break; case 15: printf("F");break; default: printf("%d",a[i]);break; } printf("\n"); }
实验任务5
方法1:不用二维数组(简)
#include<stdio.h> int main(){ int i,j,n; while(scanf("%d",&n)!=EOF){ for(j=1;j<=n;j++) {for(i=1;i<=n;i++) if(i>j) printf("%d ",j); else printf("%d ",i); printf("\n"); } }return 0; }
方法二用二维数组(复杂一点点)
#include<stdio.h> int main(){ int i,j,n; while(scanf("%d",&n)!=EOF){ int a[n][n]; for(i=1;i<=n;i++){ for(j=1;j<=n;j++) if(j>i){ a[i-1][j-1]=i; printf("%d ",a[i-1][j-1]); } else{ a[i-1][j-1]=j; printf("%d ",a[i-1][j-1]);} printf("\n"); } return 0; }}