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Re:糊涂了……哪位帮帮忙……In Reply To:糊涂了……哪位帮帮忙…… Posted by:wxn0000 at 2005-05-09 22:25:16 置换表+重排的加密不是那么容易破解的吧?肯定要做频率对应的,而且要在统计完后才做的
> 注释里的代码就可以AC
> 这样就不行了……咋回事啊?
> 没注释掉的是把msg2先按第一种加密方法处理一下
>
> #include <iostream>
> #include <string>
> using namespace std;
>
> void main()
> {
> string a, b;
> cin >> a >> b;
> int m[26] = {0,}, n[26] = {0,};
> int i;
>
>
> for( i = 0; i < a.size(); i ++ )//以下有问题
> m[a[i] - 'A'] ++;
> for( i = 0; i < b.size(); i ++ )
> {
> if( b[i] != 'Z' )
> n[b[i] - 'A' + 1] ++;
> else n[0] ++;
> }
>
> for( i = 0; i < 26; i ++ )
> {
> if( m[i] != n[i] )
> break;
> }
> if( i == 26 )
> cout << "YES" << endl;
> else cout << "NO" << endl;
>
> /* for( i = 0; i < a.size(); i ++ )//以下正确
> m[a[i] - 'A'] ++;
> for( i = 0; i < b.size(); i ++ )
> n[b[i] - 'A'] ++;
> for( i = 0; i < 26; i ++ )
> {
> for( int j = 0; j < 26; j ++ )
> {
> if( m[i] == n[j] )
> break;
> }
> n[j] = 0;
> if( j == 26 )
> {
> cout << "NO" << endl;
> return;
> }
> }
> if( i == 26 )
> cout << "YES" << endl;*/
> }
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