问题 C: A+B Problem II

题目描述

I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

A,B must be positive.

输入

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.

输出

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.

样例输入

2
1 2
112233445566778899 998877665544332211

样例输出

Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

/*对于此题做法有两种：
 其一，使2字符串的中的字符数字减去'0'，逐个相加大于等于10的可以使本位减10，下一位自增1，后面的处理就非常简单了；
 其二，便是读入字符串后先让各个字符减'0'，一一对应存入整形数组中;之后再相加。
    对于2种方法大致是相同的，都要从后面向前加，逢十进位，以及数组初始化均要初始为0，一边方便运算。
 下面用法一，代码如下：*/
#include<stdio.h>
#include<string.h>

int main( ) {
    int n, i,len1, len2, j, k, pi, t;
    char str1[1010], str2[1010];
    scanf("%d", &n);
    for(i = 1; i <= n; i++) {
        int a[1200] = {0};
        int flag = 0;
        scanf("%s%s", str1, str2);//以字符串形式读入
        printf("Case %d:
", i);
        len1 = strlen(str1);
        len2 = strlen(str2);
        printf("%s + %s = ", str1, str2);
        j = len1-1;
        k = len2-1;
        pi = 0;
        while(j >= 0 && k >= 0) { //开始相加
            if(a[pi] + (str1[j]-'0') + (str2[j]-'0') >= 10) { //相加后大于10的
                a[pi] = a[pi] + (str1[j] -'0')+(str2[k] - '0')-10;
                a[pi+1]++;
            }
            else
                a[pi] = a[pi] + (str1[j] - '0')+(str2[k]-'0');
            pi++; k--; j--;
        }
        if(j>=0) {
            for(t = j; t >= 0; t--) {
                a[pi] = a[pi]+(str1[t]-'0');
                pi ++;
            }
        }
        else if(k >= 0) {
            for(t = k; t >= 0; t--) {
                a[pi] = a[pi] + str2[t]-'0';
                pi++;
            }
        }
        else if(a[pi]!=0)//对于位数相同2个数加后最高位大于10的
            pi++;
        for(t = pi - 1; t >= 0; t--) {
            if(a[t]==0&&flag==0)
                continue;
            else
            {
                flag=1;
                printf("%d", a[t]);
            }
        }
        if(i!=n)//对于2组之间加空行的情况
            printf("
");
    }
    return 0;
}