题解 | #字符串合并处理#

//注意排序双重循环，单循环漏排；查表法实现替换。
#include <stdio.h>
#include <string.h>

int main() {
    char str1[101]={0};
    char str2[101]={0};
    char str[101]={0};
    char input[] = {"0123456789abcdefABCDEF"};
    char output[] = {"084C2A6E195D3B7F5D3B7F"};
    char tmp;
    int len = 0;
    scanf( "%s", str1 );
    scanf( "%s", str2 );
    strcat(str,str1);
    strcat(str,str2);
    len = strlen(str);
    for(int i=0; i<len; i=i+2)
    {
        for(int j=i+2; j<len; j += 2)
        {
            if(str[i] > str[j])
            {
                tmp = str[i];
                str[i] = str[j];
                str[j] = tmp;
            }
        }
       
    }
    //注意双重循环排序，单循环会漏排
    for(int i=1; i<len; i=i+2)
    {
        for(int j=i+2; j<len; j += 2)
        {
            if(str[i] > str[j])
            {
                tmp = str[i];
                str[i] = str[j];
                str[j] = tmp;
            }
        }
       
    }
    //查表实现转换
    char outstr[202] = {'\0'};
    for( int i=0; i<len; i++ )
    {
        outstr[i] = str[i];
        for( int j=0; j<22; j++ )
        {
            if( str[i] == input[j] )
                outstr[i] = output[j]; 
        }
    }

    printf("%s\n", outstr);
    return 0;
}