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  • HDU 1002 大数A+B

                  A + B Problem II



     

    Problem Description
    I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
     

     

    Input
    The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.
     

     

    Output
    For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.
     

     

    Sample Input
    2
    1 2
    112233445566778899 998877665544332211
     

     

    Sample Output
    Case 1:
    1 + 2 = 3
    Case 2:
    112233445566778899 + 998877665544332211 = 1111111111111111110
     
    代码:
     1 #include<cstring>
     2 #include<cstdio>
     3 using namespace std;
     4 
     5 int main()
     6 {
     7     //freopen("in.txt","r",stdin);
     8     int count,temp;
     9     int a1[1010],a2[1010];
    10     char s1[1010],s2[1010];
    11     int t,i,j,len1,len2;
    12     scanf("%d",&count);
    13     t=1;
    14     while(count--)
    15     {
    16         printf("Case %d:
    ",t++);
    17         memset(a1,0,sizeof(a1));
    18         memset(a2,0,sizeof(a2));
    19         scanf("%s%s",s1,s2);
    20         printf("%s + %s = ",s1,s2);
    21         int len1=strlen(s1);
    22         int len2=strlen(s2);
    23         for(i=len1-1,j=0;i>=0;i--)
    24         a1[j++]=s1[i]-'0';
    25         for(i=len2-1,j=0;i>=0;i--)
    26         a2[j++]=s2[i]-'0';
    27         int maxlen=len1>len2?len1:len2;
    28         for(i=0;i<maxlen;i++)
    29         {
    30             a1[i]+=a2[i];
    31             if(a1[i]>=10)
    32             {
    33                 a1[i+1]+=1;
    34                 a1[i]-=10;
    35             }
    36         }
    37         for(j=1009;j>=0;j--)
    38         if(a1[j])break;
    39         if(j==-1)
    40         printf("0");
    41         for(;j>=0;j--)
    42         printf("%d",a1[j]);
    43         printf("
    ");
    44         if(count!=0)
    45         printf("
    ");
    46     }
    47     return 0;
    48 }
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  • 原文地址:https://www.cnblogs.com/homura/p/4672761.html
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