hdu 1002 A + B Problem II

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A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 345393 Accepted Submission(s): 67000

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

主要分析字串一和字串二长度：短，长，相等

#include<stdio.h>
#include<string.h>
int main()
{
    int n,a,b,i,j,k;
    char str1[1005],str2[1005],str[1005];
    scanf("%d",&n);
    for(int c=1;c<=n;c++){
        memset(str1,0,sizeof(char)*1005);
        memset(str2,0,sizeof(char)*1005);
        memset(str,0,sizeof(char)*1005);
        scanf("%s",str1);
        scanf("%s",str2);
        a=strlen(str1);
        b=strlen(str2);
        i=(a>b?a:b)-1;
        j=(a>b?b:a)-1;
        int num=0;
        for(;i>=0;i--){
            if(a<=0) num+=str2[--b]-'0';
            else if(b<=0) num+=str1[--a]-'0';
            else num+=(str1[--a]-'0')+(str2[--b]-'0');
            str[i]=num%10+'0';
            num/=10;
        }
        if(num==0) printf("Case %d:\n%s + %s = %s\n",c,str1,str2,str);
        else printf("Case %d:\n%s + %s = 1%s\n",c,str1,str2,str);
        if(c!=n) printf("\n");
    }
    return 0;
}