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Question 193.2022 Winter Holiday Ladder Competition Training-7-4 Sum of N Numbers (20 points)
1. The topic
2. Problem solving
There are two key points in this question, general and approximate. Algorithms for finding the greatest common divisor and the least common multiple are required.
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll up[101],down[101];
ll gcd(ll x,ll y)//Stein算法，说是效率最高的求两数最大公约数的算法
{
x=abs(x);y=abs(y);//这个算法没法算负数
ll i,j,t;
if (x == 0 ) return y;
if (y == 0 ) return x;
for (i = 0 ; 0 == (x & 1 ); x >>= 1, ++ i);
for (j = 0 ; 0 == (y & 1 ); y >>= 1, ++ j);
if (j < i) i = j;
for (;;)
{
if (x < y) t = y,y = x,x = t;
if ( 0 == (x -= y)) return y << i;
for (; 0 == (x & 1); x >>= 1 );
}
}
/*
ll gcd(ll a, ll b){ //欧几里得算法(辗转相除)求a,b最大公约数
if(b) return gcd(b, a%b);
return a;
}
*/
ll lcm(ll x,ll y)
{
return x*y/gcd(x,y);
}
int main()
{
int N;
cin>>N;
//downlcm，upsum分别表示当前循环下的分子总和与分母的最大公约数。进入循环后不断往最终的分母最大公约数，分子总和发展
ll downlcm=1,upsum=0;//当时这里写了个int，然后我就被困了一个多小时，我真的服了我自己了
for(int i=0;i<N;i++)//采用一边输入新的分数，一边计算分数和
{
scanf("%lld/%lld",&up[i],&down[i]);
ll downtmp=downlcm;//还有这里
downlcm=lcm(down[i],downlcm);
upsum=upsum*(downlcm/downtmp)+up[i]*(downlcm/down[i]);//通分相加
}
ll resgcd=gcd(upsum,downlcm);//求出最后结果分数的分子分母最大公约数用于约分
//约分
ll res_up=upsum/resgcd;
ll res_down=downlcm/resgcd;
ll integer=res_up/res_down;
res_up=res_up%res_down;
if(integer==0&&res_up!=0)//整数为0且分子不为0则只输出分数部分
{
printf("%lld/%lld",res_up,res_down);
}
else if(res_up==0)//整数不为0且分子为0则只输出整数部分
{
printf("%lld",integer);
}
else
{
printf("%lld %lld/%lld",integer,res_up,res_down);
}
}
/*
5
-1/2 -1/2 -1/2 -1/2 -1/2
-2 -1/2（貌似1/-2也行）
*/