实验四捏
Task1_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"); /*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 不是，4字节
2 是 ，1字节
3 一样，一样
Task1_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; /*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; }
1 是 4个
2 是 1个
Task 2
#include <stdio.h> int days_of_year(int year, int month, int day); /*函数声明*/ int main() { int year, month, day; int days; while(scanf("%d%d%d", &year, &month, &day) != EOF) /*按下Ctrl+Z终止*/ { days = days_of_year(year, month, day); /*函数调用*/ 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 sum; if(year%4==0&&year%100!=0||year%400==0) { if(month==1) sum=day; if(month==2) sum=31+day; if(month==3) sum=31+29+day; if(month==4) sum=31+29+31+day; if(month==5) sum=31+29+31+30+day; if(month==6) sum=31+29+31+30+31+day; if(month==7) sum=31+29+31+30+31+30+day; if(month==8) sum=31+29+31+30+31+30+31+day; if(month==9) sum=31+29+31+30+31+30+31+30+day; if(month==10) sum=31+29+31+30+31+30+31+30+30+day; if(month==11) sum=31+29+31+30+31+30+31+30+30+31+day; if(month==12) sum=31+29+31+30+31+30+31+30+30+31+30+day; } else { if(month==1) sum=day; if(month==2) sum=31+day; if(month==3) sum=31+28+day; if(month==4) sum=31+28+31+day; if(month==5) sum=31+28+31+30+day; if(month==6) sum=31+28+31+30+31+day; if(month==7) sum=31+28+31+30+31+30+day; if(month==8) sum=31+28+31+30+31+30+31+day; if(month==9) sum=31+28+31+30+31+30+31+30+day; if(month==10) sum=31+28+31+30+31+30+31+30+30+day; if(month==11) sum=31+28+31+30+31+30+31+30+30+31+day; if(month==12) sum=31+28+31+30+31+30+31+30+30+31+30+day; } return sum; }
Task 3
#include <stdio.h> #define N 5 /*函数声明*/ 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; } /*函数定义*/ /* 输入n个整数保存到整型数组x中*/ void input(int x[], int n) { int i; for(i=0; i<n; ++i) scanf("%d", &x[i]); } /*输出整型数组x中n个元素*/ void output(int x[], int n) { int a; for(a=0; a<n; ++a) printf("%d ", x[a]); printf("\n"); } /*计算整型数组x中n个元素均值，并返回*/ /*补足函数average()实现*/ double average(int x[], int n){ double sum=0; double ave=0; int i; for(i=0;i<n;i++) sum=sum+x[i]; ave=sum/N; return ave; } /*对整型数组x中的n个元素降序排序*/ /* 补足函数sort()实现*/ void sort(int x[], int n) { int i,t,j; for(i=0;i<n-1;i++) { for(j=0;j<n-1-i;j++) { if(x[j]<x[j+1]) { t=x[j]; x[j]=x[j+1]; x[j+1]=t; } } } }
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 i; int a[20]; for(i=0;x!=0;i++) { a[i]=x%n; x=x/n; } for(i=i-1;i>=0;i--) { if(a[i]<10) { printf("%d",a[i]); } if(a[i]>=10) { if(a[i]==10) printf("A"); if(a[i]==11) printf("B"); if(a[i]==12) printf("C"); if(a[i]==13) printf("D"); if(a[i]==14) printf("E"); if(a[i]==15) printf("F"); } } printf("\n"); }
Task 5
#include<stdio.h> int main() { int a[100][100]; int n; int i, j, k; scanf("%d", &n); for(i=0;i<n;++i) { for(k=0;k<i;k++) { a[i][k]=k+1; } for(j=i;j<n;j++) { a[i][j]=i+1; } } for(i=0;i<n;i++) { for(j=0;j<n;j++) { printf("%d ",a[i][j]); } printf("\n"); } printf("\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."; printf("original\n"); puts(views1); puts(views2); char temp[N]; int i; for(i=0;i<N;i++) { temp[i]=views1[i]; } for(i=0;i<N;i++) { views1[i]=views2[i]; } for(i=0;i<N;i++) { views2[i]=temp[i]; } printf("swapping\n"); puts(views1); puts(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 t[10]; for(i=0;i<n-1;i++) for(j=0;j<n-i-1;j++) if (strcmp(str[j], str[j + 1]) > 0) { strcpy(t, str[j]); strcpy(str[j], str[j + 1]); strcpy(str[j + 1], t); } }
