poj_3979_ProblemA_分数加减法.cpp

Problem Links:

poj3979,

Problem:

分数加减法

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5903		Accepted: 1884

Description

编写一个C程序，实现两个分数的加减法

Input

输入包含多行数据
每行数据是一个字符串，格式是"a/boc/d"。

其中a, b, c, d是一个0-9的整数。o是运算符"+"或者"-"。

数据以EOF结束
输入数据保证合法

Output

对于输入数据的每一行输出两个分数的运算结果。
注意结果应符合书写习惯，没有多余的符号、分子、分母，并且化简至最简分数

Sample Input

1/8+3/8
1/4-1/2
1/3-1/3

Sample Output

1/2
-1/4
0

Source

Solution:

PS: Take care of each situation that could happen.

1st, if the numerator is 0;

2nd, if they have gcd greater than 1 after the operation.

Source Code:

//Mon Nov 22 23:32:14 CST 2010
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

using namespace std;

int gcd(int a, int b)
{
    if (b == 0)
        return a;
    else
        return gcd(b, a % b);
}

int main(int argc, const char* argv[])
{
//  freopen("input.in", "r", stdin);
//  freopen("output.out", "w", stdout);
    int a, b, c, d;
    char ch1, ch2;
    while (cin >> a >> ch1 >> b >> ch2 >> c >> ch1 >> d)
    {
        //cout << a << b << c << d << endl;

        int temp = b / gcd(b, d) * d;
        int flag = 1;
        switch (ch2)
        {
        case '+':
            a = a * (temp / b) + c * (temp / d);
            break;
        case '-':
            a = a * (temp / b) - c * (temp / d);
            flag = a < 0 ? -1 : 1;
            a = a < 0 ? -a : a;
        }
        int temp2 = gcd(a, temp);
        //cout << temp2 << endl;
        a /= temp2;
        temp /= temp2;
        if(a == 0)  //One WA here w/o considering if a is 0;
        {
            cout << "0" << endl;
            continue;
        }
        if(temp == 1)
        {
            cout << a*flag << endl; //One WA here w/o *flag;
            continue;
        }
        cout << a * flag << "/" << temp << endl;
    }
//  fclose(stdin);
//  fclose(stdout);
    return 0;
}