task4
#define _CRT_SECURE_NO_WARNINGS #include<stdio.h> #define N 10 typedef struct { char isbn[20]; char name[80]; char author[80]; double sales_price; int sales_count; }Book; void output(Book x[], int n); void sort(Book x[], int n); double sales_amount(Book x[], int n); int main() { Book x[N] = {{"978-7-229-14156-1", "源泉", "安.兰德", 84, 59}, {"978-7-5133-5261-1", "李白来到旧金山", "谭夏阳", 48, 16}, {"978-7-5617-4347-8", "陌生人日记", "周怡芳", 72.6, 27}, {"978-7-5722-5475-8", "芯片简史", "汪波", 74.9, 49}, {"978-7-5046-9568-0", "数据化决策", "道格拉斯·W·哈伯德", 49, 42}, {"978-7-5133-4388-6", "美好时代的背后", "凯瑟琳.布", 34.5, 39}, {"978-7-1155-0509-5", "无穷的开始:世界进步的本源", "戴维·多伊奇", 37.5, 55}, {"978-7-5321-5691-7", "何为良好生活", "陈嘉映", 29.5 , 31}, {"978-7-5133-5109-6", "你好外星人", "英国未来出版集团", 118, 42}, {"978-7-2011-4617-1", "世界尽头的咖啡馆", "约翰·史崔勒基", 22.5, 44}}; printf("图书销量排名:\n"); sort(x, N); output(x, N); printf("\n图书销售总额:%.2f\n", sales_amount(x, N)); return 0; } void output(Book x[], int n) { Book* p; p = x; printf("ISBN号 书名 作者 售价 销售册数\n"); while (p<x+N) { printf("%s %-30s %-30s %-10.1lf %-10d\n", p->isbn, p->name, p->author, p->sales_price, p->sales_count); p++; } } void sort(Book x[], int n) { Book temp; int i, j; for(i=0;i<n-1;i++) for(j=0;j<n-1-i;j++) if(x[j].sales_count<x[j+1].sales_count) { temp = x[j]; x[j] = x[j + 1]; x[j + 1] = temp; } } double sales_amount(Book x[], int n) { double sum = 0; int i = 0; for (i = 0; i < n; i++) { sum += x[i].sales_price* x[i].sales_count; } return sum; }
task5
#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
typedef struct {
int year;
int month;
int day;
}Date;
const int common[12] = { 31,28,31,30,31,30,31,31,30,31,30,31 };
const int leap[12] = { 31,29,31,30,31,30,31,31,30,31,30,31 };
void input(Date* pd);
int day_of_year(Date d);
int compare_dates(Date d1, Date d2);
void test1() {
Date d;
int i;
printf("输入日期\n");
for (i = 0; i < 3; ++i) {
input(&d);
printf("%d-%02d-%02d是这一年中第%d天\n\n", d.year, d.month, d.day, day_of_year(d));
}
}
void test2() {
Date Alice_birth, Bob_birth;
int i;
int ans;
printf("输入Alice和Bob出生日期:(以形如2023-12-11这样的形式输入)\n");
for (i = 0; i < 3; ++i) {
input(&Alice_birth);
input(&Bob_birth);
ans = compare_dates(Alice_birth, Bob_birth);
if (ans == 0)
printf("Alice和Bob一样大\n\n");
else if (ans == -1)
printf("Alice比Bob大\n\n");
else
printf("Alice比Bob小\n\n");
}
}
int main() {
printf("测试1: 输入日期, 打印输出这是一年中第多少天\n");
test1();
printf("\n测试2: 两个人年龄大小关系\n");
test2();
}
void input(Date* pd)
{
scanf("%d-%d-%d", &pd->year, &pd->month, &pd->day);
}
int day_of_year(Date d) {
int i = 0;
int day=0;
if ((0 == d.year % 4 && d.year % 100 != 0) || (0 == d.year % 400))
{
for (i = 0; i < d.month - 1; i++)
day += leap[i];
day += d.day;
return day;
}
else
{
for (i = 0; i < d.month - 1; i++)
day += common[i];/*在这儿,已修改*/
day += d.day;
return day;
}
}
int compare_dates(Date d1, Date d2) {
if (d1.year < d2.year)
return -1;
else if (d1.year > d2.year)
return 1;
else
{
if (d1.month < d2.month)
return -1;
else if (d1.month > d2.month)
return 1;
else
{
if (d1.day < d2.day)
return -1;
else if (d1.day > d2.day)
return 1;
else
{
return 0;
}
}
}
}
/*第一行输出错误是误把common打成leap了*/
task6
#define _CRT_SECURE_NO_WARNINGS #include <stdio.h> #include <string.h> enum Role { admin, student, teacher }; typedef struct { char username[20]; // 用户名 char password[20]; // 密码 char type[20]; // 账户类型 } Account; // 函数声明 void output(Account x[], int n); // 输出账户数组x中n个账户信息,其中,密码用*替代显示 int main() { Account x[] = { {"A1001", "123456", "student"}, {"A1002", "123abcdef","student" }, {"A1009", "xyz12121", "student"}, {"X1009", "9213071x", "admin"}, {"C11553", "129dfg32k", "teacher"}, {"X3005", "921kfmg917"," student"} }; int n; n = sizeof(x) / sizeof(Account); output(x, n); return 0; } void output(Account x[], int n) { int i = 0, j = 0,count = 0; char xin[100] = { '*','*','*','*', '*','*', '*','*', '*','*', '*','*', '*','*', '*','*', '*','*', '*','*', '*','*', '*','*', '*','*', '*','*', '*','*', '*','*','*' }; for (j = 0; j < n; j++) { count = 0; for (i = 0; i < 20; i++) { if (x[j].password[i] >= '0' && x[j].password[i] <= '9') count++; } printf("%-30s", x[j].username); for (i = 0; i < count; i++) printf("%c", xin[i]); printf(" %s\n", x[j].type); } }
标签:printf,return,int,31,编程,枚举,实验,year,day From: https://www.cnblogs.com/yrx0415/p/17894916.html