hdu 1002 A + B Problem II

偏执的太偏执、 2022-06-17 13:16 146阅读 0赞

# A + B Problem II #

**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 356395 Accepted Submission(s): 69127**

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
       
        
       
        
    
       
        
       
        
        
         #include<cstdio>
    #include<cstring>
    #include<iostream>
    using namespace std;
    #define M 100000
    void Add(char s1[],char s2[])
    {
        char num1[M]= {0},num2[M]= {0};
        int i,j,len1,len2;
        len1=strlen(s1);
        len2=strlen(s2);
        for(i=len1-1,j=0; i>=0; i--)
            num1[j++]=s1[i]-'0';
        for(i=len2-1,j=0; i>=0; i--)
            num2[j++]=s2[i]-'0';
        for(i=0; i<M; i++)
            num1[i]+=num2[i];
        for(i=0; i<M; i++)
        {
            if(num1[i]>=10)
            {
                num1[i]-=10;
                num1[i+1]++;
            }
        }
        for(i=M-1; i>=0&&num1[i]==0; i--);
        for(; i>=0; i--)
        {
            printf("%c",num1[i]+'0');
            if(i==0) printf("\n");
        }
    }
    int main()
    {
        char s1[M],s2[M];
        int n,cnt=0;
        scanf("%d",&n);
        while(n--)
        {
            cnt++;
            scanf("%s %s",s1,s2);
            printf("Case %d:\n",cnt);
            printf("%s + %s = ",s1,s2);
            Add(s1,s2);
            if(n!=0)
            printf("\n");
        }
        return 0;
    }