HDOJ 1002 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

分析

大整数相加问题。另外输出的格式需要注意

#include 
int main()
{
    int n;
    int no=0;
    char a[1001],b[1001];
    scanf("%d",&n);
    while(n--)
    {
        if(no!=0)
            printf("\n");
        no++;
        scanf("%s %s",&a,&b);
        int alength=0;
        while(a[alength]!='\0')
            alength++;
        int blength=0;
        while(b[blength]!='\0')
            blength++;
        int length=alength;
        if(length>blength)
            length=blength;
        int ans[1001];
        for(int i=0;i<1001;i++)
            ans[i]=0;
        for(int i=0;i9)
            {
                ans[i]=(t1+t2+ans[i])%10;
                ans[i+1]=1;
            }
            else
            {
                ans[i]=t1+t2+ans[i];
            }
        }
        if(length9)
                {
                    ans[i]=(a[alength-1-i]-'0'+ans[i])%10;
                    ans[i+1]=1;
                }
                else
                {
                    ans[i]=a[alength-1-i]-'0'+ans[i];
                }
            }
        }
        if(length9)
                {
                    ans[i]=(b[blength-1-i]-'0'+ans[i])%10;
                    ans[i+1]=1;
                }
                else
                {
                    ans[i]=b[blength-1-i]-'0'+ans[i];
                }
            }
        }
        printf("Case %d:\n",no);
        printf("%s + %s = ",a,b);
        int anslength=1000;
        while(ans[anslength]==0)
            anslength--;
        for(int i=anslength;i>=0;i--)
            printf("%d",ans[i]);
        printf("\n");
    }
}