pata1081（应该是long long类型却写成int类型也会超时）

最新推荐文章于 2019-05-08 21:54:00 发布

是三水不是泗水

最新推荐文章于 2019-05-08 21:54:00 发布

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本文介绍了一个算法问题：给定一系列分数形式的有理数，如何计算它们的总和并简化结果。文章通过示例展示了输入输出格式，并提供了一段C++代码实现，该算法能够处理最多100个有理数的加法运算。

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1081. Rational Sum (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Given N rational numbers in the form "numerator/denominator", you are supposed to calculate their sum.

Input Specification:

Each input file contains one test case. Each case starts with a positive integer N (<=100), followed in the next line N rational numbers "a1/b1 a2/b2 ..." where all the numerators and denominators are in the range of "long int". If there is a negative number, then the sign must appear in front of the numerator.

Output Specification:

For each test case, output the sum in the simplest form "integer numerator/denominator" where "integer" is the integer part of the sum, "numerator" < "denominator", and the numerator and the denominator have no common factor. You must output only the fractional part if the integer part is 0.

Sample Input 1:

5
2/5 4/15 1/30 -2/60 8/3

Sample Output 1:

3 1/3

Sample Input 2:

2
4/3 2/3

Sample Output 2:

#include<iostream> 
#include<vector>
#include<string>
using namespace std;  
typedef long long ll; 
ll getGCD(ll a,ll b){  
    ll temp;  
    if(a < b){  
      temp = a;  
      a = b;  
      b = temp;  
    }    
    if(a%b == 0){  
        return b;    
    }else{    
        return getGCD(b,a%b);    
    }    
}
ll doLCM(vector<ll> array,ll size){    
    ll x,y,num=array[0],i,gcd;    
    for(i=0;(i+1)<size;i++){  
        x=num;    
        y=array[i+1];    
        gcd = getGCD(x,y);      
    	num = x/gcd * y/gcd * gcd;  
    }  
    return num;    
}   

int main()
{
	ll n;
	cin>>n;
	vector<ll> va,vb;
	vector<char> vc;
	for(ll i=0;i<n;i++){
		ll a=0,b=0;
		char c;
		cin>>a>>c>>b;
		va.push_back(a);
		vb.push_back(b);
		vc.push_back(c);
	}
	ll tt=0;
	tt=doLCM(vb,vb.size());
	ll ans=0;
	for(ll i=0;i<va.size();i++){
		ans+=(va[i]*(tt/vb[i]));
	}
	if(ans==0){
		cout<<0<<endl;
	}else{
		ll k=0;
		k=getGCD(ans,tt);
		tt/=k;
		ans/=k;
		ll integer=ans/tt;
		ans-=(tt*integer);
		if(integer==0){
			if(ans!=0)
			cout<<ans<<'/'<<tt<<endl;
			else{
				cout<<0<<endl;
			}
		}else{
			if(ans!=0)
			cout<<integer<<' '<<ans<<'/'<<tt<<endl;
			else{
				cout<<integer<<endl;
			}
		}
	}
	 return 0;
}