*浙大PAT甲级 1081

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题目不难，但是需要注意的细节很多，尤其是 long long虽然数的范围在int内，但是求最小公倍数时，相乘会超出int范围，因此用long long；

AC代码：

#include<iostream>#include<map>#include<cstdio>#include<algorithm>#include<queue>#include<cstring>#include<list>#include<set>#include<stack>#include<cmath>#include<vector>#define inf 999999999using namespace std;long long gcd(long long x,long long y){    return x%y==0?y:gcd(y,x%y);}long long f(long x,long y){    return x*y/gcd(x,y);}int main(){    long long n;    scanf("%d",&n);    long long fenzi=0;    long long fenmu=1;    //cout<<fenzi<<" "<<fenmu<<endl;    for(int i=0;i<n;i++)    {        string s;        cin>>s;        long long tmp1=0;        long long tmp2=0;        int flag1=0;         if(s[0]=='-')       {        flag1=1;        s.erase(0,1);        }        int j;         for(j=0;j<s.size();j++)        {        if(s[j]=='/')            break;        tmp1*=10;        tmp1+=s[j]-'0';        }        tmp1=flag1==1?-tmp1:tmp1;        for(int r=j+1;r<s.size();r++)        {        tmp2*=10;        tmp2+=s[r]-'0';        }        long long gg=f(fenmu,tmp2);        //cout<<gg<<endl;        fenzi=fenzi*gg/fenmu;        tmp1=tmp1*gg/tmp2;        fenmu=gg;        fenzi=fenzi+tmp1;        if(fenzi==0)        {            fenmu=1;            continue;        }        long long fac=gcd(fenmu,fenzi);        //cout<<fac<<endl;        fenmu=fenmu/fac;        fenzi=fenzi/fac;        //cout<<fenzi<<" "<<fenmu<<endl;    }    if(fenmu==1)    {        if(fenzi<0)        {            printf("-%lld",-fenzi);        }        else        {            printf("%lld",fenzi);        }    }    else    {        if(fenzi/fenmu==0)        {            printf("%lld/%lld",fenzi,fenmu);        }        else        {            int d=fenzi/fenmu;            printf("%lld %lld/%lld",d,fenzi-d*fenmu,fenmu);        }    }}

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