字符串合并并处理

来源：互联网发布：中科数据恢复编辑：程序博客网时间：2024/05/16 06:49

详细描述：
将输入的两个字符串合并。
对合并后的字符串进行排序，要求为：下标为奇数的字符和下标为偶数的字符分别从小到大排序。
这里的下标意思是字符在字符串中的位置。
对排训后的字符串进行操作，如果字符为‘0’——‘9’或者‘A’——‘F’或者‘a’——‘f’，
则对他们所代表的16进制的数进行BIT倒序的操作，并转换为相应的大写字符。
如字符为‘4’，为0100b，则翻转后为0010b，也就是2。转换后的字符为‘2’；
如字符为‘7’，为0111b，则翻转后为1110b，也就是e。转换后的字符为大写‘E’。

举例：输入str1为"dec"，str2为"fab"，合并为“decfab”，
分别对“dca”和“efb”进行排序，排序后为“abcedf”，转换后为“5D37BF”
样例输入:
dec fab

样例输出:
5D37BF

#include<iostream>#include<string>#include<algorithm>using namespace std;int main(){    string s1, s2;    char sinChar[100], douChar[100];    char arr[16] = {'0','8','4','C','2','A','6','E','1','9','5','D','3','B','7','F'};    cin>>s1>>s2;    string s = s1 + s2;    //cout<<s<<endl;    int len = s.length();    int jishu = len/2 + len%2, oushu = len/2;    for(int i = 0; i < jishu; i++)        sinChar[i] = s[2*i];    for(int i = 0; i < oushu; i++)        douChar[i] = s[2*i+1];    sort(sinChar, sinChar + jishu);    sort(douChar, douChar + oushu);    /*    for(int i = 0; i < jishu; i++)        cout<<sinChar[i];    cout<<endl;    for(int i = 0; i < oushu; i++)        cout<<douChar[i];    cout<<endl;    */    for(int i = 0; i < jishu; i++){        char cc = sinChar[i];        if(cc >= '0' && cc <= '9') sinChar[i] = arr[cc - '0'];        else if(cc  >= 'a' && cc<= 'f') sinChar[i] = arr[ cc - 'a' + 10];        else if(cc  >= 'A' && cc<= 'F') sinChar[i] = arr[ cc - 'A' + 10];    }    for(int i = 0; i < oushu; i++){        char cc = douChar[i];        if(cc >= '0' && cc <= '9') douChar[i] = arr[cc - '0'];        else if(cc  >= 'a' && cc<= 'f') douChar[i] = arr[ cc - 'a' + 10];        else if(cc  >= 'A' && cc<= 'F') douChar[i] = arr[ cc - 'A' + 10];    }    for(int i = 0; i < oushu; i++)        cout<<sinChar[i]<<douChar[i];    if(len%2 == 1) cout<<sinChar[jishu-1];    cout<<endl;    return 0;}

#include <iostream>#include <string>#include <cctype>using namespace std;void sort( string& word, int old );void ProcessString( string& word );int main( void ){string word1, word2;cin >> word1 >> word2;word1.append( word2 );sort( word1, 0 );sort( word1, 1 );ProcessString( word1 );cout << word1 << endl;return 0;}void sort( string& word, int old ){char tmp;int nMax;for( nMax = old; nMax < word.length(); nMax += 2 );nMax -= 2;for( int i = nMax; i > old; i -= 2 ){bool flag = false;for( int j = old; j < i; j += 2 ){if( word[j] > word[j + 2] ){tmp = word[j];word[j] = word[j + 2];word[j + 2] = tmp;flag = true;}}if( !flag )break;}}void ProcessString( string& word ){char ch1, ch2, bit, bFlag;for( string::iterator it = word.begin();it != word.end(); ++it ){ch2 = 0;     //给char ch2赋初值为0bit = 1;bFlag = 1;if( isdigit( *it ) ){ch1 = *it - '0';}else if( *it >= 'a' && *it <= 'f' ){ch1 = *it - 'a' + 10;}else if( *it >= 'A' && *it <= 'F' ){ch1 = *it - 'A' + 10;}else{bFlag = 0;}if( bFlag ){ch2 |= ( ch1 & ( 1 << 0 ) ) << 3;ch2 |= ( ch1 & ( 1 << 1 ) ) << 1;ch2 |= ( ch1 & ( 1 << 2 ) ) >> 1;ch2 |= ( ch1 & ( 1 << 3 ) ) >> 3;if( ch2 < 10 ){*it = '0' + ch2;     //返回其字符}else{*it = 'A' + ch2 - 10;}}}}

0 0