HDU 1002 A + B Problem II

ACM OJ题解目录
本题网址：https://cn.vjudge.net/problem/HDU-1002

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:
112233445566778899 + 998877665544332211 = 1111111111111111110

题目大意：
　　大整数加法，长度最多为1000。
　　
分析：
　　首先，大数运算要用字符串来处理；其次，加法运算要考虑进位问题。
　　因此，采用字符串存储，利用ascall码来做运算，再加上一个进位计算，即可轻松解决。
　　当然，不会让你们失望的，我的小明同学又出错了。因为题说输出两个测试用例之间的空行，所以最后一个用例下面记得不要输出空行啦~
　　
代码如下：

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;const int maxn = 1000 + 5;
int t;
char s1[maxn], s2[maxn];void add(char *a, char *b)
{int alen = strlen(a) - 1;int blen = strlen(b) - 1;int clen = max(alen, blen) + 1;int carry = 0;char c[maxn], *res = c;memset(c, 0, maxn * sizeof(char));while(alen >= 0 && blen >= 0){res[clen] = a[alen] + b[blen] - '0' + carry;if(res[clen] <= '9') carry = 0;    //不用进位 else{    //进位 carry = 1;res[clen] -= 10;}alen--, blen--, clen--;}while(alen >= 0){ //如果a有剩余 res[clen] = a[alen] + carry;if(res[clen] <= '9') carry = 0;else{carry = 1;res[clen] -= 10;}alen--, clen--;}while(blen >= 0){    //如果b有剩余 res[clen] = b[blen] + carry;if(res[clen] <= '9') carry = 0;else{carry = 1;res[clen] -= 10;}blen--, clen--;}if(clen >= 0){   //如果还有进位 res[clen] = carry + '0';}if(res[0] == '0') res++;    //去除首位0 cout << a << " + " << b << " = " << res << endl;
}int main()
{cin >> t;for(int i = 1; i <= t; i++){memset(s1, 0, maxn * sizeof(char));memset(s2, 0, maxn * sizeof(char));cin >> s1 >> s2;cout << "Case " << i << ":" << endl;add(s1, s2);if(i < t) cout << endl;}return 0;
}

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