HDU 2100 分类: ACM 2015-06-17 23:49 15人阅读 评论(0) 收藏

Lovekey

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6721    Accepted Submission(s): 2186


Problem Description
XYZ-26进制数是一个每位都是大写字母的数字。 A、B、C、…、X、Y、Z 分别依次代表一个0 ~ 25 的数字,一个 n 位的26进制数转化成是10进制的规则如下
A0A1A2A3…An-1 的每一位代表的数字为a0a1a2a3…an-1 ,则该XYZ-26进制数的10进制值就为

m = a0 * 26^(n-1) + a1 * 26^(n-2) + … + an-3* 26^2 + an-2*26 + an-1

一天vivi忽然玩起了浪漫,要躲在学校的一个教室,让枫冰叶子去找,当然,她也知道枫冰叶子可不是路痴,于是找到了XYZ的小虾和水域浪子帮忙,他们会在vivi藏的教室的门口,分别写上一个XYZ-26进制数,分别为 a 和 b,并且在门锁上设置了密码。显然,只有找到密码才能打开锁,顺利进入教室。这组密码被XYZ的成员称为lovekey。庆幸的是,枫冰叶子知道lovekey是 a的10进制值与b的10进制值的和的XYZ-26进制形式。当然小虾和水域浪子也不想难为枫冰叶子,所以a 和 b 的位数都不会超过200位。
例如第一组测试数据
a = 0 * 26^5+0* 26^4+ 0* 26^3+ 0 *26^2 + 3*26 + 7 = 85
b = 1*26^2 + 2*26 + 4 = 732
则 a + b = 817 = BFL
 

Input
题目有多组测试数据。
每组测试数据包含两个值均为的XYZ-26进制数,每个数字的每位只包含大写字母,并且每个数字不超过200位。
 

Output
输出XYZ的lovekey,每组输出占一行。
 

Sample Input
  
    
AAAADH BCE DRW UHD D AAAAA
 

Sample Output
  
    
BFL XYZ D
 


#include<queue>
#include<math.h>
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define N 215
char a[N],b[N],c[N];
void reverse(char a[])
{
    int len=strlen(a);
    for(int i=0;i<len/2;i++)
    {
        swap(a[i],a[len-i-1]);
    }
}
int main()
{
    while(~scanf("%s%s",a,b))
    {

        int len_a=strlen(a);
        int len_b=strlen(b);
        reverse(a);
        reverse(b);
        int jin=0;
        for(int i=0;i<max(len_a,len_b);i++)
        {
//            cout<<b[i]<<endl;
            if(a[i]=='\0')
            {
//                printf("aaaaajin=%d\n",jin);
                if(jin)
                {
                    c[i]=b[i]-'A'+1;
                    jin=0;
                    if(c[i]>=26)
                    {
                        if(i==max(len_a,len_b)-1)
                        {
                            c[i]=c[i]-26+'A';
                            c[i+1]='B';
                            c[i+2]=0;
                        }
                        else
                        {
                            c[i]=c[i]-26+'A';
                            jin=1;
                        }
                    }
                    else
                    {
                        c[i]=c[i]+'A';
                    }
                }
                else
                {
                    c[i]=b[i];
                    jin=0;
                }

            }
            else if(b[i]=='\0')
            {
//                printf("bbbbbjin=%d\n",jin);
                if(jin)
                {
                    c[i]=a[i]-'A'+1;
                    jin=0;
                    if(c[i]>=26)
                    {
                        if(i==max(len_a,len_b)-1)
                        {
                            c[i]=c[i]-26+'A';
                            c[i+1]='B';
                            c[i+2]=0;
                        }
                        else
                        {
                            c[i]=c[i]-26+'A';
                            jin=1;
                        }
                    }
                    else
                    {
                        c[i]=c[i]+'A';
                    }
                }
                else
                {
                    c[i]=a[i];
                    jin=0;
                }
            }
//            printf("%c %c %c\n",a[i],b[i],jin);
            else {

//            printf("%c %c\n",a[i],b[i]);
            c[i]=a[i]+b[i]+jin-'A'*2;
//            printf(" *  %d \n",c[i]);
            jin=0;
            if(c[i]>=26)
            {
                if(i==max(len_a,len_b)-1)
                {
                    c[i]=c[i]-26+'A';
                    c[i+1]='B';
                    c[i+2]=0;
                }
                else
                {
                    c[i]=c[i]-26+'A';
                    jin=1;
//                    printf("jin=%d",jin);
                }
            }
            else
                c[i]=c[i]+'A';
//            printf(" #  %d %c\n",c[i],c[i]);
        }

        }

        for(int i=strlen(c)-1;i>0;i--)
        {
            if(c[i]=='A')
                c[i]='\0';
            else
            {
                break;
            }
        }
        reverse(c);

        cout<<c<<endl;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(c,0,sizeof(c));

    }
    return 0;
}




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