A + B Problem II

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:

思路:从末尾开始加起，但是由与两个数的长度可能不同，末尾对应的下标会不同，计算的时候不方便，我们可以先将两个数倒置再相加，如：
     6451564 +
1234567896514
倒置后变成
4651546 +
4156987654321
这样末尾的下标相同，计算起来更加方便，最后再将计算结果倒着输出即可。

#include <stdio.h>#include <string.h>char num1[10005];char num2[10005];int ans[10005];//保存计算结果 void invert(char a[], int len)//倒置功能 {    char temp;    for (int i = 0; i <= len/2-1; i++)    {        temp = a[i];        a[i] = a[len-i-1];        a[len-i-1] = temp;    }}int main(){    int t,count = 0;    scanf("%d",&t);    while (t--)    {        count++;        scanf("%s%s",num1,num2);        int len1 = strlen(num1);        int len2 = strlen(num2);        int lenmax = len1 > len2 ? len1 : len2;        int lenmin = len1 < len2 ? len1 : len2;        invert(num1,len1);//倒置两个数         invert(num2,len2);        int flag = 0;//保存进位         for (int i = 0; i < lenmax; i++)        {            if (i < lenmin)            {                ans[i] = (num1[i]+num2[i]-'0'-'0'+flag)%10;                flag = (num1[i]+num2[i]-'0'-'0'+flag)/10;            }            else            {                if (len1 > len2)                 {                    ans[i] = (num1[i]-'0'+flag)%10;                    flag = (num1[i]-'0'+flag)/10;                }                else                {                    ans[i] = (num2[i]-'0'+flag)%10;                    flag = (num2[i]-'0'+flag)/10;                }            }        }        if (flag != 0)        {            ans[lenmax] = flag;            lenmax++;        }        if (count > 1) printf("\n");        printf("Case %d:\n",count);        invert(num1,len1);//将两个数倒置回来         invert(num2,len2);        printf("%s + %s = ",num1,num2);        for (int i = lenmax-1; i >= 0; i--)        {            printf("%d",ans[i]);        }        printf("\n");    }    return 0;}