http://codeforces.com/problemset/problem/1/B 题意：给出了两种表格的表示方法。要求互相转化。 思路：直接模拟即可。注意和一般的进制转化不同的是，26进制对应的是1到26而不是0到25，所以要记得处理下借位。
代码实现
1/* ***********************************************
2Author :111qqz
3Created Time :2015年12月13日 星期日 19时46分09秒
4File Name :code/cf/problem/1B.cpp
5************************************************ */
6
7#include <cstdio>
8#include <cstring>
9#include <iostream>
10#include <algorithm>
11#include <vector>
12#include <queue>
13#include <set>
14#include <map>
15#include <string>
16#include <cmath>
17#include <cstdlib>
18#include <ctime>
19#define fst first
20#define sec second
21#define lson l,m,rt<<1
22#define rson m+1,r,rt<<1|1
23#define ms(a,x) memset(a,x,sizeof(a))
24typedef long long LL;
25#define pi pair < int ,int >
26#define MP make_pair
27
28using namespace std;
29const double eps = 1E-8;
30const int dx4[4]={1,0,0,-1};
31const int dy4[4]={0,-1,1,0};
32const int inf = 0x3f3f3f3f;
33
34char st[100];
35int a26[100];
36int a10[100];
37
38void pre()
39{
40 a26[0]=1;
41 for ( int i = 1 ;i<=5; i++)
42 {
43 a26[i] = a26[i-1]*26;
44// cout<<"i:"<<i<<" "<<a26[i]<<endl;
45 }
46 a10[0] = 1;
47 for ( int i = 1 ; i<=7 ; i++)
48 {
49 a10[i] = a10[i-1]*10;
50 }
51}
52void solve()
53{
54 cin>>st;
55 int len = strlen(st);
56 int p1=-1,p2=-1;
57 for ( int i = 0 ; i < len ; i++)
58 {
59 char ch = st[i];
60 if (ch>='A'&&ch<='Z')
61 {
62 if (p1==-1)
63 {
64 p1 = i ;
65 }
66 else
67 {
68 p2 = i;
69 break;
70 }
71 }
72 }
73 if (p2-p1==1||p2==-1)
74 {
75 int dig = 0 ;
76 int alp = 0;
77 int sum = 0 ;
78 int sum2 = 0 ;
79 for ( int i = len -1 ; i>= 0 ; i--)
80 {
81 char ch = st[i];
82 if (!isdigit(ch))
83 {
84 sum += (ch-'A'+1)*a26[alp];
85 alp++;
86 }
87 else
88 {
89 sum2+= (ch-'0')*a10[dig];
90 dig++;
91 }
92 }
93 printf("R%d\n",sum2,sum);
94 }
95 else
96 {
97 int dig1 = 0 ;
98 int dig2 = 0;
99 int sum1 = 0 ;
100 int sum2 = 0 ;
101 int p = 1;
102 for ( int i = len-1 ; i>= 0 ; i-- )
103 {
104 char ch = st[i];
105 if (isdigit(ch))
106 {
107 // cout<<"ch:"<<ch<<endl;
108 if (p)
109 {
110 sum1+=(ch-'0')*a10[dig1];
111 // cout<<"sum1:"<<sum1<<endl;
112 dig1++;
113 }
114 else
115 {
116 sum2+=(ch-'0')*a10[dig2];
117 dig2++;
118 }
119
120
121 }
122 else
123 {
124 p = 0 ;
125 }
126 }
127// cout<<"sum1:"<<sum1<<" sum2:"<<sum2<<endl;
128 int cnt = 0 ;
129 int b[50];
130 while(sum1)
131 {
132 cnt++;
133 b[cnt] = sum1;
134 sum1/=26;
135// cout<<"sum1::::"<<sum1<<endl;
136 }
137// cout<<"cnt:"<<cnt<<endl;
138 for ( int i = 1 ; i <= cnt -1 ; i ++)
139 {
140 if (b[i]<=0)
141 {
142 b[i]+=26;
143 b[i+1]--;
144 }
145 }
146 while (b[cnt]==0) cnt--;
147
148 for ( int i = cnt ; i >=1 ; i--)
149 {
150
151 // cout<<"b[i]:"<<b[i]<<endl;
152 cout<<char(b[i]+'A'-1);
153 }
154 cout<<sum2<<endl;
155
156
157
158
159 }
160
161
162}
163int main()
164{
165 #ifndef ONLINE_JUDGE
166 freopen("code/in.txt","r",stdin);
167 #endif
168
169 pre();
170 int T;
171 cin>>T;
172 while (T--)
173 {
174 solve();
175 }
176
177 #ifndef ONLINE_JUDGE
178 fclose(stdin);
179 #endif
180 return 0;
181}