1002，大数求和

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

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思路分析：
共分三步进行处理，1，两数组各自从最高下标相加并存如新数组中，直到其中一个到达下标为0；
2,将下标没到达0的数组加到新数组中；
3,判断最后是否还有进位，若有加到新数组中。

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#include<stdio.h>#include<string.h>int main(){    int t;    int i,j,k,m,l1,l2;    char a[1005],b[1005];    int c[1005];    scanf ("%d",&t);    int x=1;    while (t--)    {        scanf ("%s%s",a,b);        int m = 0;        l1 = strlen(a);        l2 = strlen(b);        for (i=l1-1,j=l2-1,k=0; i>=0 && j>=0; i--,j--)        {            m = a[i]-48+b[j]-48+m;            c[k++] = m%10;            m /= 10;        }        if (k == l1)        {            while (j>=0)            {                m = b[j--]-48+m;                c[k++] = m%10;                m = m/10;            }        }        else        {            while (i>=0)            {                m = a[i--]-48+m;                c[k++] = m%10;                m = m/10;            }        }        if (m != 0)        {            c[k++] = m;        }        printf ("Case %d:\n",x++);        printf ("%s + %s = ",a,b);        for (i=k-1; i>=0; i--)        {            printf ("%d",c[i]);        }        printf ("\n");        if (t)            printf ("\n");    }    return 0;}