问题描述

按如下函数原型用函数编程解决如下的日期转换问题(要求考虑闰年的问题)

解决方案

 #include <stdio.h>int mt[13] = { 0 31 28 31 30 31 30 31 31 30 31 30 31 };void set2(int y){    if (y % 4 != 0)    {        mt[2] = 28;    }    else    {        if (y % 100 == 0 && y % 400 != 0)            mt[2] = 28;        else            mt[2] = 29;    }}void showmenu(){    printf(""1. y/n/d -> ydn"");    printf(""2. yd -> y/n/dn"");    printf(""3. exitn"");    }int ymd2yd(int y int m int d){    set2(y);    int r = 0;    for (int i = 0; i < m; i++)        r += mt[i];    r += d;    return r;}void yd2ymd(int yd int * r){    set2(r[0]);    int m = 0;    while (yd > mt[m])    {        yd -= mt[m++];    }    r[1] = m;    r[2] = yd;}int main(){       char c;    showmenu();    while (scanf(""%c"" &c) 1)    {        switch (c)        {        case '1':            int y int m int d;            scanf(""%d"" &y);            scanf(""%d"" &m);            scanf(""%d"" &d);            printf(""%dn"" ymd2yd(y m d));            showmenu();            break;        case '2':            int yd;            int r[3];            scanf(""%d"" &r[0]);            scanf(""%d"" &yd);            yd2ymd(yd r);            printf(""%d %d %dn"" r[0] r[1] r[2]);            showmenu();            break;        case 'n':            break;        default:            return 0;        }    }    return 0;}

解决方案二：

1. y/n/d -> yd
2. yd -> y/n/d
3. exit 12016 3 161
4. y/n/d -> yd
5. yd -> y/n/d
6. exit 12016 7 18200
7. y/n/d -> yd
8. yd -> y/n/d
9. exit 12015 12 31365
10. y/n/d -> yd
11. yd -> y/n/d
12. exit 22016 612016 3 1
13. y/n/d -> yd
14. yd -> y/n/d
15. exit 22015 612015 3 2
16. y/n/d -> yd
17. yd -> y/n/d
18. exit 22015 2252015 8 13
19. y/n/d -> yd
20. yd -> y/n/d
21. exit 3Press any key to continue

解决方案三：

 1. y/n/d -> yd2. yd -> y/n/d3. exit 12016 3 11. y/n/d -> yd2. yd -> y/n/d3. exit 12016 7 181. y/n/d -> yd2. yd -> y/n/d3. exit 12015 12 311. y/n/d -> yd2. yd -> y/n/d3. exit 22016 612016 3 11. y/n/d -> yd2. yd -> y/n/d3. exit 22015 612015 3 21. y/n/d -> yd2. yd -> y/n/d3. exit 22015 2252015 8 131. y/n/d -> yd2. yd -> y/n/d3. exit 3Press any key to continue