B

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

s思路及要点

这是一道很典型的大数运算问题，我们可以利用字符串来模拟竖式运算的方法解决。注意的要点有：1、字符型数组要转化为整形数组，需要在每一位上减去‘0’（即字符型的0）。2、为了方便运算，我们需要将加数与被加数的最低位对齐，所以在将字符数组转化为整形数组时，可以将数组逆序排列，保证最低位对齐。3、注意进位。4、由于运算时数组已被逆置，所以在输出时，需要逆序输出，才能保证结果正确。5、注意输出格式，是一次换行还是两次换行。

#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;char a[1005], b[1005];int main(){    int t;    while(scanf("%d", &t) != EOF){        int kase = 0;        while(t--){            scanf("%s%s", a, b);            int numa[1005] = {0}, numb[1005] = {0}, numc[1005] = {0};            int len_a = strlen(a), len_b = strlen(b);            int ma = max(len_a, len_b);            for(int i = len_a - 1; i >= 0; i--)                numa[len_a - 1 - i] = a[i] - '0';            for(int j = len_b - 1; j >= 0; j--)                numb[len_b - 1 - j] = b[j] - '0';            for(int i = 0; i < ma; i++)                numc[i] = numa[i] + numb[i];            for(int i = 0; i < ma; i++){                    if(numc[i] >= 10){                        if(i == ma - 1)ma++;                        numc[i] -= 10;                        numc[i+1]++;                    }            }            if(kase)printf("\n");            printf("Case %d:\n%s + %s = ", ++kase, a, b);            for(int i = ma - 1; i >= 0; i--)                printf("%d", numc[i]);            printf("\n");        }    }    return 0;}