Problem solution calculates a score P1572 []

String good question! Other explanationsReadability is not highI use the most simple ideas to explain it again

analysis:

• If you read a string into a word processing from the numerator and denominator will be very troublesome, and when read with scanf("%d/%d",&a,&b)can easily deal with the sign of the words

• Common denominator: a a*b=lcm(a,b)*gcd(a,b), toolcm(a,b)=a * b/gcd(a,b)

  This question pit

• 1:00 denominator is required special sentence
• Results into the most simple integer ratio: we can count on every time when they are taken over __gcd (a, b) <--- This is a library function, hands-free play again gcd,Although much time did not save

• Most pit point: When the denominator is negative, eg 2/-3become-2/3

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define rep(i,a,b) for(ll i=a;i<=b;++i)
#define dwn(i,a,b) for(ll i=a;i>=b;--i) 
 
template <typename T> inline void rd(T &x){
    x=0;char c=getchar();int f=0;
    while(!isdigit(c)){f|=c=='-';c=getchar();}
    while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();}
    x=f?-x:x;
}
#define mem(a,b) memset(a,b,sizeof(a))

int a,b,c,d;
int ans1;//分子 
int ans2;//分母 

int main(){
    scanf("%d/%d",&a,&b);
    while(scanf("%d/%d",&c,&d)!=EOF){
        int gcd=__gcd(b,d);
        int lcm=b*d/gcd;
        ans2=lcm;//当前分母
        ans1=a*(lcm/b)+c*(lcm/d);//当前分子，请读者自主模拟一下
        a=ans1/(__gcd(ans1,ans2));//下一次计算时，a和b代表的是上一次计算的分子分母，顺便约个分
        b=ans2/(__gcd(ans1,ans2));
    }
    ans1=a;//最后一次处理后a就是ans1,这里赋值的原因是a与b约过一次分了
    ans2=b;//最后一次处理后b就是ans2
    
    if(ans2<0){//处理 2/-3 -> -2/3的情况 
        ans1=-ans1;
        ans2=-ans2;
    }
    if(ans2==1)printf("%d\n",ans1);//处理分母为1的情况
    else printf("%d/%d\n",ans1,ans2);
    return 0;
}