很水的一道题，提醒自己要认真，做的头都快晕了。考虑26的特殊情况。
D - Spreadsheets
Time Limit:10000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit Status
Description
In the popular spreadsheets systems (for example, in Excel) the following numeration of columns is used. The first column has number A, the second — number B, etc. till column 26 that is marked by Z. Then there are two-letter numbers: column 27 has number AA, 28 — AB, column 52 is marked by AZ. After ZZ there follow three-letter numbers, etc.
The rows are marked by integer numbers starting with 1. The cell name is the concatenation of the column and the row numbers. For example, BC23 is the name for the cell that is in column 55, row 23.
Sometimes another numeration system is used: RXCY, where X and Y are integer numbers, showing the column and the row numbers respectfully. For instance, R23C55 is the cell from the previous example.
Your task is to write a program that reads the given sequence of cell coordinates and produce each item written according to the rules of another numeration system.
Input
The first line of the input contains integer number n (1 ≤ n ≤ 105), the number of coordinates in the test. Then there follow n lines, each of them contains coordinates. All the coordinates are correct, there are no cells with the column and/or the row numbers larger than 106.
Output
Write n lines, each line should contain a cell coordinates in the other numeration system.
Sample Input
Input
2R23C55BC23
Output
BC23R23C55
1 #include<cstdio>
2 #include<cstring>
3 #include<cstdlib>
4 #include<cctype>
5 using namespace std;
6 const int maxn=100;
7 char str[maxn];
8 int num[maxn];
9 void print(int n)
10 {
11 if(n==1)
12 printf("A");
13 if(n==2)
14 printf("B");
15 if(n==3)
16 printf("C");
17 if(n==4)
18 printf("D");
19 if(n==5)
20 printf("E");
21 if(n==6)
22 printf("F");
23 if(n==7)
24 printf("G");
25 if(n==8)
26 printf("H");
27 if(n==9)
28 printf("I");
29 if(n==10)
30 printf("J");
31 if(n==11)
32 printf("K");
33 if(n==12)
34 printf("L");
35 if(n==13)
36 printf("M");
37 if(n==14)
38 printf("N");
39 if(n==15)
40 printf("O");
41 if(n==16)
42 printf("P");
43 if(n==17)
44 printf("Q");
45 if(n==18)
46 printf("R");
47 if(n==19)
48 printf("S");
49 if(n==20)
50 printf("T");
51 if(n==21)
52 printf("U");
53 if(n==22)
54 printf("V");
55 if(n==23)
56 printf("W");
57 if(n==24)
58 printf("X");
59 if(n==25)
60 printf("Y");
61 if(n==26||n<=0)
62 printf("Z");
63
64 }
65 int main()
66 {
67 int T;
68 scanf("%d",&T);
69 while(T--)
70 {
71 int q=0;
72 scanf("%s",str);
73 int len=strlen(str);
74 int flag=1;
75 int ok=0;
76 int sum=0;
77 int temp=0;
78 int temp2=0;
79 for(int i=0; i<len; i++)
80 {
81 if(flag==1&&isalpha(str[i])==0)
82 {
83 flag=0;
84 temp2=i;
85 }
86 if(flag==0&&isalpha(str[i])!=0)
87 {
88 ok=1;
89 temp=i;
90 break;
91 }
92 }
93 if(ok)
94 {
95 int p=1;
96 for(int i=len-1; i>temp; i--)
97 {
98 sum+=(str[i]-‘0‘)*p;
99 p*=10;
100 }
101 // if(sum%26==0) 考虑错了
102 // {
103 // while(sum>0)
104 // {
105 // num[q++]=sum%26-1;
106 // if(sum==26)
107 // break;
108 // sum/=26;
109 // }
110 // }
111 // else
112
113 while(sum>0)
114 {
115 if(sum%26==0)
116 {
117 num[q++]=26;
118 sum=sum/26-1;
119 }
120 else{
121 num[q++]=sum%26;
122 sum/=26;
123 }
124 }
125 for(int i=q-1; i>=0; i--)
126 {
127 // printf("%d ",num[i]);
128 print(num[i]);
129 }
130 for(int i=1; i<temp; i++)
131 printf("%c",str[i]);
132 }
133 else
134 {
135 printf("R");
136 for(int i=temp2; i<len; i++)
137 printf("%c",str[i]);
138 printf("C");
139 int p=1;
140 for(int i=temp2-1; i>=0; i--)
141 {
142 sum+=(str[i]-‘A‘+1)*p;
143 p*=26;
144 }
145 printf("%d",sum);
146 }
147 printf("\n");
148 }
149 }