zoukankan      html  css  js  c++  java
  • Kattis之旅——Rational Arithmetic

    Input

    The first line of input contains one integer, giving the number of operations to perform.

    Then follow the operations, one per line, each of the form x1 y1 op x2 y2. Here, −109≤x1,y1,x2,y2<109 are integers, indicating that the operands are x1/y1 and x2/y2. The operator op is one of ’+’, ’−’, ’∗’, ’/’, indicating which operation to perform.

    You may assume that y1≠0, y2≠0, and that x2≠0 for division operations.

    Output

    For each operation in each test case, output the result of performing the indicated operation, in shortest terms, in the form indicated.

    Sample Input 1Sample Output 1
    4
    1 3 + 1 2
    1 3 - 1 2
    123 287 / 81 -82
    12 -3 * -1 -1
    
    5 / 6
    -1 / 6
    -82 / 189
    -4 / 1
    

    题目不解释了,看样例也能懂。

    注意用long long。

     //Asimple
    #include <bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    ll n, m, s, res, ans, len, T, k, num;
    ll a, b, c, d;
    char ch;
    
    ll gcd(ll a, ll b ) {
        return b==0?a:gcd(b, a%b);
    }
    
    void print(ll num, ll den){
        bool pos = (num>0 && den>0) || (num<0 && den<0);
        if (num<0) num = -num;
        if (den<0) den = -den;
        ll d = gcd(num,den);
        num /= d , den /= d;
        if (num==0 || den==0) cout << "0 / 1" << endl;
        else cout << (pos?"":"-") << num << " / " << den << endl;
    }
    
    void add_sub(ll x1, ll y1, ll x2, ll y2, int state){
        ll num = x1*y2 + state*x2*y1;
        ll den = y1*y2;
        print(num,den);
    }
    
    void mul(ll x1, ll y1, ll x2, ll y2){
        ll num = x1*x2;
        ll den = y1*y2;
        print(num,den);
    }
    
    void input() {
        cin >> T;
        while( T -- ) {
            cin >> a >> b >> ch >> c >> d;
            switch(ch){
                case '+':
                    add_sub(a, b, c, d,1);
                    break;
                case '-':
                    add_sub(a, b, c, d, -1);
                    break;
                case '*':
                    mul(a, b, c, d);
                    break;
                case '/':
                    mul(a, b, d, c);
                    break;
            }
        }
    }
    
    int main(){
        input();
        return 0;
    }
  • 相关阅读:
    如何在SpringMVC中使用REST风格的url
    c#实现的udt
    数据库查询服务化-缓存和分页
    c#常用数据库封装再次升级
    c#数据库连接池Hikari重构升级
    c# 常用数据库封装
    聊聊数据存储查询
    c#分析SQL语句
    c# 分析SQL语句中的表操作
    c#最近博文结尾
  • 原文地址:https://www.cnblogs.com/Asimple/p/6773648.html
Copyright © 2011-2022 走看看