任务一：颠倒的世界
小明最近突然喜欢倒着写字，写出来的句子全是颠倒的，也就是把一句话里的字符全都逆序写，譬如“I Love This Game！”，他就偏偏要写成“!emaG sihT evoL I”弄的小华同他Email交流都很难受。请你帮助小华解决这个问题，将输入的字符串全都还原成本来的面目。
任务1要求：
利用main函数带参数的功能，编写程序，将命令行中的字符串全部逆序显示出来。例如编写的可执行文件为go.exe 那么运行 go !emaG sihT evoL I 时，显示出I Love This Game！
任务二：时间转换
由于工作原因，小明常常需要签署文件日期，但他比较偷懒，都用阿拉伯数字签署，譬如3.8表示三月八日，7.12表示七月十二日。最近公司改组成外资企业，主管要求将所有的日期全部更改为英文表示，一来同国际接轨，二来也不容易篡改，譬如将3.8表示为March the eighth 面对这么多的日期，小明犯愁了，他英文不好，那么多的英文单词他记不住。而且小明还是个马大哈，有时候居然会签出3.32（三月三十二日）这样的日期，为了避免被上司责备，趁着这次机会也赶快一并修改。请你帮助小明完成这个任务。
任务2要求：
利用main函数带参数的功能，编写程序，将命令行中的字符串翻译成英文表示，每行显示一个日期，同时需检查日期表示是否有误，如有误，请给出提示。例如：编写的可执行文件为go.exe 那么运行go 3.8 5.5 4.38 时显示
March the eighth
May the fifth
Error
任务三：时间转换(续)
小明觉得这么一点点的手工输入还是有点麻烦，他把需要翻译的日期都整理到文本文件in.txt中，请你帮助小明，将其中的内容翻译完毕后保存到文本文件out.txt中，便于他统一处理。
任务3要求：翻译过程同任务2，将结果保存至文件out.txt中
第一题：
1 #include<stdio.h> 2 #include<stdlib.h> 3 #include<string.h> 4 #define max(a,b) ((a)>(b)?(a):(b)) 5 #define min(a,b) ((a)<(b)?(a):(b)) 6 #define abs(a) ((a)>0?(a):(-(a))) 7 #define sqr(a) ((a)*(a)) 8 #define swap(a,b) (a)^=(b),(b)^=(a),(a)^=(b) 9 #define eps 1e-8 10 #define MAX 0x7f7f7f7f 11 #define N 104 12 int a[N]; 13 int main(int argc,char* argv[]) 14 { 15 int i,j,k; 16 for(i=argc-1;i>0;i--) 17 { 18 for(j=strlen(argv[i])-1;j>=0;j--) 19 putchar(argv[i][j]); 20 printf(" "); 21 } 22 puts(""); 23 return 0; 24 } 25 /* 26 Q:xxx111222inDebug222 !emaG sihT evoL I 27 */
第二题：
1 #include<stdio.h> 2 #include<stdlib.h> 3 #include<string.h> 4 #define max(a,b) ((a)>(b)?(a):(b)) 5 #define min(a,b) ((a)<(b)?(a):(b)) 6 #define abs(a) ((a)>0?(a):(-(a))) 7 #define sqr(a) ((a)*(a)) 8 #define swap(a,b) (a)^=(b),(b)^=(a),(a)^=(b) 9 #define eps 1e-8 10 #define MAX 0x7f7f7f7f 11 #define N 104 12 int m,d; 13 int mon[]={0,31,28,31,30,31,30,31,31,30,31,30,31}; 14 char *M[]={"","January","February","March","April","May","June","July","August","September","October","November","December"}; 15 char *D[]={"","first","second","third","fourth","fifth","sixth","seventh","eighth","ninth","tenth","eleventh","twelfth","thirteenth","fourteenth","fifteenth","sixteenth","seventeenth","eighteenth","nineteenth","twentieth","twenty-first","twenty-second","twenty-third","twenty-fourth","twenty-fifth","twenty-sixth","twenty-seventh","twenty-eighth","twenty-ninth","thirtieth","thirty-first"}; 16 void work(char s[]) 17 { 18 int i; 19 m=d=0; 20 if(s[2]=='.') 21 { 22 m=(s[0]-'0')*10+s[1]-'0'; 23 i=3; 24 } 25 else if(s[1]=='.') 26 { 27 m=s[0]-'0'; 28 i=2; 29 } 30 for(;i<strlen(s);i++) 31 d=d*10+s[i]-'0'; 32 } 33 int main(int argc,char * argv[]) 34 { 35 int i,j,k; 36 for(i=1;i<argc;i++) 37 { 38 work(argv[i]); 39 if(d>mon[m] || d<1 || m>12 ||m<1)puts("Error"); 40 else printf("%s the %s",*(M+m),*(D+d)); 41 puts(""); 42 } 43 return 0; 44 } 45 /* 46 Q:xxx111222inDebug222 3.8 5.5 4.38 47 */
第三题：
1 #include<stdio.h> 2 #include<stdlib.h> 3 #include<string.h> 4 #define max(a,b) ((a)>(b)?(a):(b)) 5 #define min(a,b) ((a)<(b)?(a):(b)) 6 #define abs(a) ((a)>0?(a):(-(a))) 7 #define sqr(a) ((a)*(a)) 8 #define swap(a,b) (a)^=(b),(b)^=(a),(a)^=(b) 9 #define eps 1e-8 10 #define MAX 0x7f7f7f7f 11 #define N 1004 12 int m,d; 13 int mon[]={0,31,28,31,30,31,30,31,31,30,31,30,31}; 14 char argv[N]; 15 char *M[]={"","January","February","March","April","May","June","July","August","September","October","November","December"}; 16 char *D[]={"","first","second","third","fourth","fifth","sixth","seventh","eighth","ninth","tenth","eleventh","twelfth","thirteenth","fourteenth","fifteenth","sixteenth","seventeenth","eighteenth","nineteenth","twentieth","twenty-first","twenty-second","twenty-third","twenty-fourth","twenty-fifth","twenty-sixth","twenty-seventh","twenty-eighth","twenty-ninth","thirtieth","thirty-first"}; 17 void work(char s[]) 18 { 19 int i; 20 m=d=0; 21 if(s[2]=='.') 22 { 23 m=(s[0]-'0')*10+s[1]-'0'; 24 i=3; 25 } 26 else if(s[1]=='.') 27 { 28 m=s[0]-'0'; 29 i=2; 30 } 31 for(;i<strlen(s);i++) 32 d=d*10+s[i]-'0'; 33 } 34 int main() 35 { 36 FILE *fp1=fopen("Q:\xxx\111\111\in.txt","r"),*fp=fopen("Q:\xxx\111\111\out.txt","w"); 37 int i,j,k; 38 while(~fscanf(fp1,"%s",argv)) 39 { 40 work(argv); 41 if(d>mon[m] || d<1 || m>12 ||m<1)fputs("Error ",fp); 42 else fprintf(fp,"%s the %s ",*(M+m),*(D+d)); 43 } 44 fclose(fp1); 45 fclose(fp); 46 return 0; 47 } 48 /* 49 Q:xxx111111inDebug111 3.8 5.5 4.38 13.1 1.32 1.0 2.29 115123 3.31 50 */