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.

1234567890
abcdefghij
99999jjjjj
9999900001

Sample Output:

bdfi02467j
iiiij00000


題目意思：二十進制加法運算



。。。。我的代碼：

 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這種情況,0不能輸出，將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 }

大佬的代碼是這樣的：

 1 #include<bits/stdc++.h>///學習：可以將要輸出的內容寫入一個字符數組之中，輸出在字符數組中的位置即可
 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 }

學習了

Martian Addition