Base64

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description

Mike does not want others to view his messages, so he find a encode method Base64.

Here is an example of the note in Chinese Passport.

The Ministry of Foreign Affairs of the People‘s Republic of China requests all civil and military authorities of foreign countries to allow the bearer of this passport to pass freely and afford assistance in case of need.

When encoded by \texttt{Base64}, it looks as follows

VGhlIE1pbmlzdHJ5IG9mIEZvcmVpZ24gQWZmYWlycyBvZiB0aGUgUGVvcGxlJ3MgUmVwdWJsaWMgb2Yg

Q2hpbmEgcmVxdWVzdHMgYWxsIGNpdmlsIGFuZCBtaWxpdGFyeSBhdXRob3JpdGllcyBvZiBmb3JlaWdu

IGNvdW50cmllcyB0byBhbGxvdyB0aGUgYmVhcmVyIG9mIHRoaXMgcGFzc3BvcnQgdG8gcGFzcyBmcmVl

bHkgYW5kIGFmZm9yZCBhc3Npc3RhbmNlIGluIGNhc2Ugb2YgbmVlZC4=

In the above text, the encoded result of \texttt{The} is \texttt{VGhl}. Encoded in ASCII, the characters \texttt{T}, \texttt{h}, and \texttt{e} are stored as the bytes84,104,
and 101,
which are the 8-bit
binary values 01010100,01101000,
and 01100101.
These three values are joined together into a 24-bit string, producing
010101000110100001100101.

Groups of 6
bits (6
bits have a maximum of 26=64
different binary values) are converted into individual numbers from left to right (in this case, there are four numbers in a 24-bit string), which are then converted into their corresponding Base64 encoded characters. The Base64 index table is

0123456789012345678901234567890123456789012345678901234567890123

ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/

In the above example, the string 010101000110100001100101
is divided into four parts 010101,000110,100001
and 100101,
and converted into integers 21,6,33
and 37.
Then we find them in the table, and get V, G, h, l.

When the number of bytes to encode is not divisible by three (that is, if there are only one or two bytes of input for the last 24-bit block), then the following action is performed:

Add extra bytes with value zero so there are three bytes, and perform the conversion to base64. If there was only one significant input byte, only the first two base64 digits are picked (12 bits), and if there were two significant input bytes, the first three
base64 digits are picked (18 bits). ‘=‘ characters are added to make the last block contain four base64 characters.

As a result, when the last group contains one bytes, the four least significant bits of the final 6-bit block are set to zero; and when the last group contains two bytes, the two least significant bits of the final 6-bit block are set to zero.

For example, base64(A) = QQ==, base64(AA) = QUE=.

Now, Mike want you to help him encode a string for
k
times. Can you help him?

For example, when we encode A for two times, we will get base64(base64(A)) = UVE9PQ==.

Input

The first line contains an integer
T(T≤20)
denoting the number of test cases.

In the following T
lines, each line contains a case. In each case, there is a number
k(1≤k≤5)
and a string s.s
only contains characters whose ASCII value are from
33
to 126(all
visible characters). The length of s
is no larger than 100.

Output

For each test case, output Case #t:, to represent this is t-th case. And then output the encoded string.

Sample Input

2
1 Mike
4 Mike

Sample Output

Case #1: TWlrZQ==
Case #2: Vmtaa2MyTnNjRkpRVkRBOQ==

题意：将3个8位变成4个6位，不足八位的时候补0，最后不足四位补=，求出k次变换后的结果

#include <bits/stdc++.h>
using namespace std;
const int N = 10005;
char e[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";

//int main()
//{
//    int k=11;
//    for(int i=7;i>=0;i--)
//    cout<<(k>>i&1);
//    cout<<endl;
//    //00001011
//
//    k=(11<<8);
//
//    for(int i=15;i>=0;i--)
//    cout<<(k>>i&1);
//    cout<<endl;
//    //0000101100000000
//
//    return 0;
//}

void base64(char s[])
{
    char t[N]={0};//要加一个对数组的初始化。
    int l=strlen(s);
    int p=0;
    int a=0;

    int r=l%3;

    for(int i=0;i<l;i+=3)
    {
        int k=(((int)s[i])<<16)+(((int)s[i+1])<<8)+(int)(s[i+2]);//每次取三个字节
        int cb=1;

        for(int j=0;j<24;j++)
        {
            a+=(k>>j&1)*cb;
            cb<<=1;
            if(j%6==5)
            {
                t[p++]=e[a];
                a=0;
                cb=1;
            }
        }
        swap(t[p-1],t[p-4]);//4位字符是反的，比如AAA （100000101000001010000010）每6位一组后是BFUQ  倒过来后是QUFB 正常AAA应该是（010000010100000101000001）从左到右六个一组就是QUFB

        swap(t[p-2],t[p-3]);
    }

    if(r==1)
    t[p-1]=t[p-2]='=';
    if(r==2)
    t[p-1]='=';

    memcpy(s,t,sizeof(char)*N);
}

int main()
{
    int cas, n;
    scanf("%d", &cas);
    char s[N];
    for(int k = 1; k <= cas; ++k)
    {
        memset(s,0,sizeof(s));
        scanf("%d%s", &n, s);
        while(n--)
            base64(s);
        printf("Case #%d: %s\n", k, s);
    }
    return 0;
}
/*
0123456789012345678901234567890123456789012345678901234567890123
ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/
*/