园龄：6年11个月粉丝：2052 关注：178

Martian Addition

In the 22nd Century, scientists have discovered intelligent residents live on the Mars. Martians are very fond of mathematics. Every year, they would hold an Arithmetic Contest on Mars (ACM). The task of the contest is to calculate the sum of two 100-digit numbers, and the winner is the one who uses least time. This year they also invite people on Earth to join the contest.

As the only delegate of Earth, you're sent to Mars to demonstrate the power of mankind. Fortunately you have taken your laptop computer with you which can help you do the job quickly. Now the remaining problem is only to write a short program to calculate the sum of 2 given numbers. However, before you begin to program, you remember that the Martians use a 20-based number system as they usually have 20 fingers.

Input:
You're given several pairs of Martian numbers, each number on a line.
Martian number consists of digits from 0 to 9, and lower case letters from a to j (lower case letters starting from a to present 10, 11, ..., 19).
The length of the given number is never greater than 100.

Output:
For each pair of numbers, write the sum of the 2 numbers in a single line.

Sample Input:

1234567890
abcdefghij
99999jjjjj
9999900001

Sample Output:

bdfi02467j
iiiij00000


题目意思：二十进制加法运算



。。。。我的代码：

 1 #include<stdio.h>
 2 #include<string.h>
 3 int a[200],b[200],c[200];
 4 char x[200],s[200],t[200];
 5 int main()
 6 {
 7     int i,j,k,len1,len2,len;
 8     while(scanf("%s%s",x,s)!=EOF)
 9     {
10         memset(a,0,sizeof(a));
11         memset(b,0,sizeof(b));
12         memset(c,0,sizeof(c));
13         len1=strlen(x);
14         len2=strlen(s);
15         if(len1>len2)
16             len=len1;
17         else
18             len=len2;
19         i=0;
20         for(j=len1-1; j>=0; j--)
21         {
22             if(x[j]<='9'&&x[j]>='0')
23                 a[i++]=x[j]-'0';
24             else
25                 a[i++]=x[j]-87;
26         }
27         i=0;
28         for(j=len2-1; j>=0; j--)
29         {
30             if(s[j]<='9'&&s[j]>='0')
31                 b[i++]=s[j]-'0';
32             else
33                 b[i++]=s[j]-87;
34         }
35         for(i=0; i<len; i++)
36         {
37             c[i]=c[i]+a[i]+b[i];
38             if(c[i]>19)
39             {
40                 c[i+1]=c[i]/20;
41                 c[i]=c[i]%20;
42             }
43         }
44         i=len;
45         while(!c[i])///000001这种情况,0不能输出，将0跳出
46         {
47             i--;
48             if(i==-1)
49             {
50                 printf("0");
51                 break;
52             }
53         }
54         k=0;
55         for(j=i; j>=0; j--)
56         {
57             if(c[j]>=0&&c[j]<=9)
58                 t[k++]=c[j]+'0';
59             else
60                 t[k++]=c[j]+87;
61         }
62         for(j=0; j<k; j++)
63             printf("%c",t[j]);
64         printf("\n");
65 
66     }
67     return 0;
68 }

大佬的代码是这样的：

 1 #include<bits/stdc++.h>///学习：可以将要输出的内容写入一个字符数组之中，输出在字符数组中的位置即可
 2 int main()
 3 {
 4     char a[200],b[200],s[21]="0123456789abcdefghij";
 5     int c[200],i,j,k,p,q;
 6     while(scanf("%s%s",a,b)!=EOF){
 7         memset(c,0,sizeof(c));
 8         k=0;
 9         p=strlen(a);
10         q=strlen(b);
11         j=q-1;
12         for(i=p-1;i>=0||j>=0;i--){
13             if(i>=0){
14                 if(a[i]-'a'>=0)
15                     c[k]+=a[i]-'a'+10;
16                 else
17                     c[k]+=a[i]-'0';
18             }
19             if(j>=0){
20                 if(b[j]-'a'>=0)
21                     c[k]+=b[j]-'a'+10;
22                 else
23                     c[k]+=b[j]-'0';
24             }
25             if(c[k]>=20){
26                 c[k]%=20;
27                 c[k+1]+=1;
28             }
29             j--;
30             k++;
31         }
32         if(c[k]==0)
33             k-=1;
34         for(i=k;i>=0;i--)
35             printf("%c",s[c[i]]);
36         printf("\n");
37     }
38     return 0;
39 }