NYOJ 303

View Code

 1 #include<iostream>

 2 #include<cstdio>

 3 #include<cstdlib>

 4 #include<cstring>

 5 #include<stack>

 6 using namespace std;

 7 

 8 char ch[1000];

 9 stack <int> q;

10 

11 void slovenum()

12 {

13   int i,len;

14   while(!q.empty())q.pop();

15   int x=0;

16   len=strlen(ch);

17   for(i=0;i<len;++i)

18    x=x*10+ch[i]-'0';

19   while(x)

20    {

21      q.push(x%26);

22      if(x==26)break;

23      if(x%26==0)x-=26;//注意这里：当 x = 52时即余数为 0 由于只有26个字母，不能当27进制处理 

24      else x-=x%26;

25      x/=26;

26    }

27   while(!q.empty())

28    {

29      int temp=q.top();

30      q.pop();

31      if(!temp)printf("Z");

32      else 

33      printf("%c",temp-1+'A');

34    }

35   printf("\n");

36 }

37 

38 void slovechar()

39 {

40   int i,len;

41   len=strlen(ch);

42   int sum=0;

43   for(i=0;i<len;++i)

44    sum=sum*26+ch[i]-'A'+1;

45   printf("%d\n",sum);

46 }

47 

48 int main()

49 {

50   int iscase,i;

51   scanf("%d",&iscase);

52   while(iscase--)

53    {

54      scanf("%s",ch);

55      if(ch[0]>='0'&&ch[0]<='9')

56       slovenum();

57      else slovechar();

58    }

59   return 0;

60 }