HDU 5237-Base64（模拟-K轮加密）

Base64

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1242    Accepted Submission(s): 556

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 bytes 84, 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 21,6,33and 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

21 Mike4 Mike

 

Sample Output

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

 

Source

The 2015 ACM-ICPC China Shanghai Metropolitan Programming Contest

 

题目意思：

T组测试数据，每组K轮加密S字符串。

先根据每个字符的ASCII码转换成八位二进制，然后分割成六位二进制码转换回ASCII，如果末尾不足六位需要补零，如果当前轮加密后的字符串长度不是四的倍数，需要在末尾补‘#’。

解题思路：

模拟…注意处理好细节就好啦~

#include <iostream>#include <cstdio>#include <cstring>#include <bitset>#include <map>#include <iomanip>#include <algorithm>#define MAXN 100000#define INF 0xfffffffusing namespace std;char c[MAXN];int a[MAXN];string str[MAXN];char ans[MAXN];char b[MAXN];char f[65]= {'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z','a','b','c','d','e','f','g','h','i','j',             'k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z','0','1','2','3','4','5','6','7','8','9','+','/'            };//加密规则void solve(char c[]){    memset(a,'\0',sizeof(a));//存储ASCII码    memset(b,'\0',sizeof(b));//存储二进制串    memset(ans,'\0',sizeof(ans));    int len=strlen(c);    int cnt=0,cans=0;    for(int i=0; i<len; ++i)    {        a[i]=int(c[i]);        bitset<8>s(a[i]);//转换成八位二进制串        string c=s.to_string<char,char_traits<char>,allocator<char> >();        for(int j=0; j<8; ++j) b[cnt++]=c[j];    }    int l=0,r=0,lenb;    if(cnt%6!=0) l=cnt,r=((cnt/6)+1)*6;//计算需要末尾补零时    bool bl=false;//标志是否进行补零操作    if(r!=0&&l!=0)    {        bl=true;        for(int i=l; i<r; ++i)            b[i]='0';//补零        lenb=r;//更新b数组长度    }    else lenb=cnt;    l=0,r=0;    for(int i=0; i<cnt; ++i)    {        if(i!=0&&(i+1)%6==0)//每六位一组计算        {            r=i;//当前六位数组的起始下标            char ch[10];            int cn=0;            for(int j=l; j<=r; ++j)                ch[cn++]=b[j];            string str2(ch);            bitset<6> bs2(str2);//将六位二进制数转换成ASCII码            int res=bs2.to_ulong();            ans[cans++]=f[res];            l=i+1;//当前六位数组的结束下标        }    }    if(bl)//计算补零后最后六位    {        char ch[10]= {'\0'};        int cn=0;        for(int i=l; i<lenb; ++i)            ch[cn++]=b[i];        string str2(ch);        bitset<6> bs2(str2);        int res=bs2.to_ulong();        ans[cans++]=f[res];//加密后的字符存入数组    }    if(cans%4!=0) l=cans,r=((cans/4)+1)*4;//计算需要末尾补‘#’时    for(int i=l; i<r; ++i)        ans[cans++]='=';    /* for(int i=0; i<cans; ++i)         cout<<ans[i];     cout<<endl;*/    memset(c,'\0',sizeof(c));    for(int i=0; i<cans; ++i)//本次计算得到的加密串复制到原串c以便下次加密        c[i]=ans[i];}int main(){#ifdef ONLINE_JUDGE#else    freopen("G:/cbx/read.txt","r",stdin);    //freopen("G:/cbx/out.txt","w",stdout);#endif    ios::sync_with_stdio(false);    cin.tie(0);    int T,ca=0;    cin>>T;    while(T--)    {        int k;        cin>>k;        memset(c,'\0',sizeof(c));        cin>>c;        for(int i=0; i<k; ++i)//k次加密            solve(c);        int len=strlen(c);        cout<<"Case #"<<++ca<<": ";        for(int i=0; i<len; ++i)//输出结果            cout<<c[i];        cout<<endl;    }    return 0;}