HDU 1002 A + B Problem II

http://acm.hdu.edu.cn/showproblem.php?pid=1002

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 277135    Accepted Submission(s): 53489

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

Author

Ignatius.L

Recommend

比较简单的字符串处理，直接上代码吧！！！

附上AC代码：

#include <iostream>
#include <cstdio>
#include <string>
#include <cmath>
#include <iomanip>
#include <ctime>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <deque>
//#pragma comment(linker, "/STACK:102400000, 102400000")
using namespace std;
typedef long long ll;
const double pi = acos(-1.0);
const double e = exp(1.0);
const double eps = 1e-8;
const int maxlen = 1005;
char a[maxlen], b[maxlen], sum[maxlen];

int main(){
	ios::sync_with_stdio(false);
	int T, cas=0;
	scanf("%d", &T);
	while (T--){
		scanf("%s%s", a, b);
		int len1 = strlen(a)-1;
		int len2 = strlen(b)-1;
		int len = (len1 > len2 ? len1 : len2);
		bool flag = false;
		memset(sum, 0, sizeof(sum));
		while (len1>=0 && len2>=0){
			int num = int(a[len1]+b[len2]-'0'-'0');
			if (flag) num++;
			sum[len] = char(num%10+'0');
			len1--; len2--; len--;
			if (num >= 10)
				flag = true;
			else
				flag = false;
		}
		cas++;
		printf("Case %d:\n%s + %s = ", cas, a, b);
		if (len1 >= 0){
			while (len1 >= 0){
				int num = int(a[len1]-'0');
				if (flag) num++;
				sum[len] = char(num%10+'0');
				len1--; len--;
				if (num >= 10)
					flag = true;
				else
					flag = false;
			}
		}
		else if (len2 >= 0){
			while (len2 >= 0){
				int num = int(b[len2]-'0');
				if (flag) num++;
				sum[len] = char(num%10+'0');
				len2--; len--;
				if (num >= 10)
					flag = true;
				else
					flag = false;
			}
		}
		if (flag)
			printf("1");
		printf("%s\n", sum);
		if (T != 0)
			printf("\n");
	}
	return 0;
}