简单模拟题。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<stack>
#include<queue>
#include<string>
#include<algorithm>
using namespace std; struct FenShu
{
long long fz,fm;
FenShu(long long a,long long b)
{
fz=a;
fm=b;
}
}; long long gcd(long long a,long long b)
{
if(b==) return a;
return gcd(b,a%b);
} FenShu ADD(FenShu a,FenShu b)
{
FenShu res(,);
res.fz=a.fz*b.fm+b.fz*a.fm;
res.fm=a.fm*b.fm; if(res.fz!=)
{
long long GCD=gcd(abs(res.fz),abs(res.fm));
res.fz=res.fz/GCD;
res.fm=res.fm/GCD;
}
else
{
res.fz=;
res.fm=;
}
return res;
} int main()
{
int n; scanf("%d",&n);
FenShu ans(,);
for(int i=;i<=n;i++)
{
long long fz,fm; scanf("%lld/%lld",&fz,&fm);
FenShu t(fz,fm);
ans=ADD(ans,t);
}
//printf("%d/%d\n",ans.fz,ans.fm);
if(ans.fz==) printf("0\n");
else
{
if(ans.fz%ans.fm==) printf("%lld\n",ans.fz/ans.fm);
else if(abs(ans.fz)<ans.fm) printf("%lld/%lld\n",ans.fz,ans.fm);
else
{
long long d=ans.fz/ans.fm;
ans.fz=ans.fz-d*ans.fm;
printf("%lld %lld/%lld\n",d,ans.fz,ans.fm);
}
}
return ;
}

PAT (Advanced Level) 1081. Rational Sum (20)的更多相关文章

  1. 【PAT甲级】1081 Rational Sum (20 分)

    题意: 输入一个正整数N(<=100),接着输入N个由两个整数和一个/组成的分数.输出N个分数的和. AAAAAccepted code: #define HAVE_STRUCT_TIMESPE ...

  2. PAT (Advanced Level) 1088. Rational Arithmetic (20)

    简单题. 注意:读入的分数可能不是最简的.输出时也需要转换成最简. #include<cstdio> #include<cstring> #include<cmath&g ...

  3. PAT (Advanced Level) Practice 1008 Elevator (20 分) 凌宸1642

    PAT (Advanced Level) Practice 1008 Elevator (20 分) 凌宸1642 题目描述: The highest building in our city has ...

  4. PAT (Advanced Level) Practice 1035 Password (20 分) 凌宸1642

    PAT (Advanced Level) Practice 1035 Password (20 分) 凌宸1642 题目描述: To prepare for PAT, the judge someti ...

  5. PAT Advanced 1081 Rational Sum (20) [数学问题-分数的四则运算]

    题目 Given N rational numbers in the form "numerator/denominator", you are supposed to calcu ...

  6. PAT甲题题解-1081. Rational Sum (20)-模拟分数计算

    模拟计算一些分数的和,结果以带分数的形式输出注意一些细节即可 #include <iostream> #include <cstdio> #include <algori ...

  7. 1081. Rational Sum (20)

    the problem is from PAT,which website is http://pat.zju.edu.cn/contests/pat-a-practise/1081 the code ...

  8. 1081. Rational Sum (20) -最大公约数

    题目如下: Given N rational numbers in the form "numerator/denominator", you are supposed to ca ...

  9. PAT (Advanced Level) Practice 1035 Password (20 分)

    To prepare for PAT, the judge sometimes has to generate random passwords for the users. The problem ...

随机推荐

  1. HDU 5965 三维dp 或 递推

    题意:= =中文题 思路一:比赛时队友想的...然后我赛后想了一下想了个2维dp,但是在转移的时候,貌似出了点小问题...吧?然后就按照队友的思路又写了一遍. 定义dp[i][j][k],表示第i列, ...

  2. list组件

    <?xml version="1.0"?> <!-- Simple example to demonstrate the Spark List component ...

  3. php发送get、post请求获取内容的几种方法

    方法1: 用file_get_contents 以get方式获取内容 <?php $url='http://www.domain.com/'; $html = file_get_contents ...

  4. getopt(分析命令行参数)

    ref:http://vopit.blog.51cto.com/2400931/440453   相关函数表头文件         #include<unistd.h>定义函数       ...

  5. Array.length vs Array.prototype.length

    I found that both the Array Object and Array.prototype have the length property. I am confused on us ...

  6. ignite中的sql查询

    ignite中进行sql查询需要对要查询的cache和字段进行配置,可以在xml中配置,也可以在代码中配置或进行注解,我用的是xml配置: <!-- 配置cache --> <pro ...

  7. jquery获得select的文本

    本来以为jQuery("#select1").val();是取得选中的值, 那么jQuery("#select1").text();就是取得的文本. 这是不正确 ...

  8. 代码创建xml文档并写入指定节点

    //首先创建 XmlDocument xml文档 XmlDocument xml = new XmlDocument(); //创建根节点 config XmlElement config = xml ...

  9. linux 用户管理维护 清缓存

    #echo 1 > /proc/sys/ vm/drop_caches 2013.10.10 其实一直user group一直都没去弄清楚 只是没去归类,@@一种是对用户/组直接修改(同时也更改 ...

  10. asp.net MVC 3多语言方案--再次写, 配源码

    之前写了一篇asp.net MVC多语言方案,那次其实是为American Express银行开发的.有许多都是刚开始接触,对其也不太熟悉.现在再回过头去看,自己做一个小网站,完全用asp.net m ...