Days Past Due 30+
时间: 2024-01-17 08:04:33 浏览: 7
Days past due 30 refers to a payment that is 30 days late. This means that the payment was due on a specific date, but the payment was not made until 30 days after that date. This could result in late fees or penalties, and could negatively impact the credit score of the person or company that is responsible for making the payment. It is important to make payments on time to avoid these consequences.
相关问题
#include <stdio.h> #include <stdlib.h> struct Date { int year; int month; int day; }; int days(struct Date date) { int num = 0; int two = 0; if (date.year % 4 == 0) { if (date.year % 100 != 0) { two = 29; } else { two = 28; } if (date.year % 400 == 0) { two = 29; } } if (date.year % 400 == 0) { two = 29; } switch (date.month) { case 1: num = 0; break; case 2: num = 31; break; case 3: num = 31 + two; break; case 4: num = 31+31 + two; break; case 5: num = 30+ 31 + 31 + two; break; case 6: num = 31+ 30 + 31 + 31 + two; break; case 7: num = 30+ 31 + 30 + 31 + 31 + two; break; case 8: num = 31+ 30 + 31 + 30 + 31 + 31 + two; break; case 9: num = 31+ 31 + 30 + 31 + 30 + 31 + 31 + two; break; case 10:num = 30+ 31 + 31 + 30 + 31 + 30 + 31 + 31 + two; break; case 11: num = 31+ 30 + 31 + 31 + 30 + 31 + 30 + 31 + 31 + two; break; case 12: num = 30+ 31 + 30 + 31 + 31 + 30 + 31 + 30 + 31 + 31 + two; break; } return num + date.day; } int main() { struct Date a; scanf("%d%d%d", &a.year, &a.month, &a.day); printf("%d %d %d", a.year, a.month, a.day); printf("这一天是本年中的%d", cal(a)); system("pause"); return 0; }
这是一个计算给定日期是该年的第几天的程序。它首先定义了一个结构体Date表示日期,然后定义了一个函数days来计算输入的日期是该年的第几天。在days函数中,它使用了闰年的判断规则来计算二月份的天数,并通过switch语句来计算给定日期之前的所有天数。最后,它在main函数中读取输入的日期,并输出该日期及其在该年中的天数。
long operator int ()const //成员转换函数 { long ndays = 0; //步骤1) for (int i = 1; i < yr; ++i) //步骤2) ndays += Leapyear(i) ? 366 : 365; for (int i = 1; i < mo; ++i) //步骤3) ndays += dys[i - 1]; if (mo > 2 && Leapyear(yr)) //闰年闰月加一天 ++ndays; ndays += day; //步骤4) return ndays; 怎么改进为正确形式
这段代码需要的改进是将其转换为正确的形式。目前这段代码存在以下问题:
1. 不应该将 `operator int()` 定义为 `const` 成员函数,因为它会修改对象的状态。
2. 在计算年份对应的天数时,应该考虑当前年份是否为闰年。如果当前年份是闰年,应该将该年份的天数加一。
3. 在计算月份对应的天数时,应该考虑当前月份之前的所有月份的天数总和。
下面是改进后的代码:
```
operator int() {
int ndays = 0;
for (int i = 1; i < yr; ++i) {
ndays += Leapyear(i) ? 366 : 365;
}
if (Leapyear(yr) && mo > 2) {
ndays += 1; // 当前年份是闰年且当前月份大于2,因此需要加一天
}
for (int i = 1; i < mo; ++i) {
ndays += dys[i - 1];
}
ndays += day;
return ndays;
}
```