1081. Rational Sum (20)

1.每次读取一个数时,对其进行因式化简,再累加

2.每次直接进行相加,分子先不对分母取模,在最后才进行取模,才把整数部分提取出来

3.注意在计算gcd时,取bot和top的绝对值,避免算出来的gcd为-1,使得top和bot的符号翻转

4.刚开始尝试使用scanf("%d/%d",&a,&b)进行分式的读取,结果在读取负数分式的时候出错,所以改为读取string,然后再进行解析


//#include<string>
//#include <iomanip>
//#include<stack>
//#include<unordered_set>
//#include <sstream>
//#include "func.h"
//#include <list>
#include<unordered_map>
#include<set>
#include<queue>
#include<map>
#include<vector>
#include <algorithm>
#include<stdio.h>
#include<iostream>
#include<string>
#include<memory.h>
#include<limits.h>
#include<stack>
using namespace std;
/*
3
1/3 -1/6 1/8

3
0/1 0/2 0/3

3
-1/2 -3/2 -1/3

4
-1/2 -3/2 -1/3 1/3

4
1/2 3/2 -1/3 1/3

4
1/2 3/2 -1/2 -3/2
*/
long long gcd(long long a, long long b)
{
	return (b == 0 ? a : gcd(b, a%b));
}
struct ratNode{
	long long top;
	long long bot;
	ratNode(int a, int b) :top(a), bot(b){};
	ratNode() :top(0), bot(1){};

};
void add2Num(long long&nowTop, long long&nowBot, long long nextTop, long long nextBot)
{
	long long botGCD = gcd(nowBot, nextBot);
	long long a = nextBot / botGCD;
	long long b = nowBot / botGCD;
	long long top = nowTop*a + nextTop*b;
	long long bot = a*nowBot;
	long long GCD = gcd(labs(top), labs(bot));//注意取绝对值,避免算出来的gcd为-1,使得top和bot的符号翻转
	nowTop = top / GCD;
	nowBot = bot / GCD;
}
void str2num(string str, long long&top, long long&bot)
{
	bool sign = true;
	bool first = true;
	top = 0; bot = 0;
	for (int i = 0; i < str.size(); i++)
	{
		if (str[i] == '-')
			sign = false;
		else if (str[i] == '/')
			first = false;
		else if (first)
			top = top * 10 + str[i] - '0';
		else if (!first)
			bot = bot * 10 + str[i] - '0';
	}
	if (!sign) top = -top;
}
int main(void)
{
	int n;
	cin >> n;
	vector<ratNode> rat(n);
	long long nowTop = 0;
	long long nowBot = 1;
	for (int i = 0; i < n; i++)
	{
		string str;
		cin >> str;
		str2num(str, rat[i].top, rat[i].bot);
		if (rat[i].top == 0) rat[i].bot = 1;
		int GCD = gcd(labs(rat[i].top), labs(rat[i].bot));//注意取绝对值,避免算出来的gcd为-1,使得top和bot的符号翻转
		rat[i].top /= GCD;
		rat[i].bot /= GCD;
		add2Num(nowTop, nowBot, rat[i].top, rat[i].bot);
	}

	long long nowInt = nowTop / nowBot;
	nowTop = nowTop - (nowInt*nowBot);
	if (nowTop < 0 && nowInt!=0 ) nowTop = -nowTop;

	if (nowTop == 0 && nowInt==0)
		cout << "0" << endl;
	else if (nowTop == 0)
		cout << nowInt << endl;
	else if (nowInt == 0)
		cout << nowTop << "/" << nowBot << endl;
	else 
		cout << nowInt << " " << nowTop << "/" << nowBot << endl;

	return 0;
}


posted @ 2015-11-24 12:49  siukwan  阅读(148)  评论(0)    收藏  举报