nyoj 103 A + B problem II

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A+B Problem II

时间限制：3000 ms  |  内存限制：65535 KB

难度：3

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

大数加法，以前写的，可读性很差,唯一有用的地方是reverse函数，用来逆置一个字符串

 
#include<string.h>
#include<stdio.h>
#include<iostream>
#include <algorithm>
using namespace std;
char str[2][1002];
int main(){
	int len[2];
	int num;
	bool flag;
	int i , j;
	scanf("%d", &num);
	for(j= 1; j <=num; j++)	{
		memset(str, 0, sizeof(str));
		scanf("%s %s", str[0], str[1]);
		printf("Case %d:\n", j);
		printf("%s + %s = " , str[0] , str[1]);
		len[0] = strlen(str[0]);
		len[1] = strlen(str[1]);
		reverse(str[0], str[0] + len[0]);
		reverse(str[1], str[1] + len[1]);
		if(len[0] > len[1])	flag = 0;
		else flag = 1;
		for(i= 0 ; str[!flag][i]; i++){
			str[flag][i] += str[!flag][i] - '0';
		}
		for(i = 0 ;str[flag][i]; i++){
			if(str[flag][i] > '9'){
				if(str[flag][i + 1] < '0')
					str[flag][i + 1] = '1';
				else
					str[flag][i + 1] ++;
				str[flag][i] -= 10;
			}
		}
		len[flag] = strlen(str[flag]);
		reverse(str[flag], str[flag] + len[flag]);
		printf("%s\n" , str[flag]);
	}
	return 0;
}