不妨说，管理咋不和谐你？
by 自动Wrong机 @ 2017-02-08 19:37:17
虽然与题目有关联
by 自动Wrong机 @ 2017-02-08 19:38:12
#有种你发题解
~~装X~~
by Malgré_L_Guerre @ 2017-02-08 19:49:51
这表可以速生成吧……
如果在NOIP当中需要用这类表，我会迅速打开Python生成。
下面给大家一个生成这个表的Python程序(和上面的表不太一样，但大体相同)
```python
l=[]
for i in range(ord('A'),ord('A')+26):
s='"'
for j in range(i,ord('Z')+1):
s+=chr(j)
for j in range(ord('A'),i):
s+=chr(j)
l.append(s+'"')
print("string x[]={"+",\n".join(l)+"};")
```
by Sweetlemon @ 2017-02-08 20:36:19
[查看楼上代码及运行结果](http://ideone.com/fYWuOE)
by Sweetlemon @ 2017-02-08 20:38:18
ac代码：
#include <bits/stdc++.h>
using namespace std;
string x[30] = {"\*ABCDEFGHIJKLMNOPQRSTUVWXYZ",
"AABCDEFGHIJKLMNOPQRSTUVWXYZ",
"BBCDEFGHIJKLMNOPQRSTUVWXYZA",
"CCDEFGHIJKLMNOPQRSTUVWXYZAB",
"DDEFGHIJKLMNOPQRSTUVWXYZABC",
"EEFGHIJKLMNOPQRSTUVWXYZABCD",
"FFGHIJKLMNOPQRSTUVWXYZABCDE",
"GGHIJKLMNOPQRSTUVWXYZABCDEF",
"HHIJKLMNOPQRSTUVWXYZABCDEFG",
"IIJKLMNOPQRSTUVWXYZABCDEFGH",
"JJKLMNOPQRSTUVWXYZABCDEFGHI",
"KKLMNOPQRSTUVWXYZABCDEFGHIJ",
"LLMNOPQRSTUVWXYZABCDEFGHIJK",
"MMNOPQRSTUVWXYZABCDEFGHIJKL",
"NNOPQRSTUVWXYZABCDEFGHIJKLM",
"OOPQRSTUVWXYZABCDEFGHIJKLMN",
"PPQRSTUVWXYZABCDEFGHIJKLMNO",
"QQRSTUVWXYZABCDEFGHIJKLMNOP",
"RRSTUVWXYZABCDEFGHIJKLMNOPQ",
"SSTUVWXYZABCDEFGHIJKLMNOPQR",
"TTUVWXYZABCDEFGHIJKLMNOPQRS",
"UUVWXYZABCDEFGHIJKLMNOPQRST",
"VVWXYZABCDEFGHIJKLMNOPQRSTU",
"WWXYZABCDEFGHIJKLMNOPQRSTUV",
"XXYZABCDEFGHIJKLMNOPQRSTUVW",
"YYZABCDEFGHIJKLMNOPQRSTUVWX",
```cpp
"ZZABCDEFGHIJKLMNOPQRSTUVWXY"};
inline char findchar(char n1, char n2){
int k1, k2, i;
for (i = 1; i <= 26; ++ i){
if (x[0][i] == n1){
k1 = i;
break;
}
}
for (i = 1; i <= 26; ++ i){
if (x[i][k1] == n2){
k2 = i;
break;
}
}
return x[k2][0];
}
int main(){
int i;
vector<int> jishu;
string miyaok, miwen, ans;
getline(cin, miyaok), getline(cin, miwen);
for (i = 0; i < miwen.size(); ++ i)
if (miwen[i] >= 'a' && miwen[i] <= 'z') jishu.push_back(i);
for (i = 0; i < miyaok.size(); ++ i) miyaok[i] = toupper(miyaok[i]);
for (i = 0; i < miwen.size(); ++ i) miwen[i] = toupper(miwen[i]);
for (i = 0; i < miwen.size(); ++ i)
ans.push_back(findchar(miyaok[i % miyaok.size()], miwen[i]));
for (i = 0; i < jishu.size(); ++ i)
ans[jishu[i]] = tolower(ans[jishu[i]]);
cout << ans << endl;
return 0;
}
```
by Alextokc @ 2017-02-09 15:32:11
@[Alextokc](/space/show?uid=26050) 可是，根本不需要这表
by 1124828077ccj @ 2017-02-09 19:13:50
@[2016陈常杰](/space/show?uid=14738) 党然啦。。。
by Alextokc @ 2017-02-09 19:39:37
HEHE
by QRcode @ 2017-02-28 20:18:44
LINUXtr，strupr都不行
by pupuvovovovovo @ 2017-05-03 20:20:14