2980. 小数转化分数

#include<bits/stdc++.h>
using namespace std;
using u64 = uint64_t;
int main()
{
    int T; cin >> T;
    for(int cs = 0; cs < T; cs++)
    {
        cout << "case #" << cs << ":\n";
        string s; cin >> s;
        auto dot = s.find('.');
        u64 a = stoull(s.substr(0, dot)), n, m;
        auto rec = s.find('[');
        if(rec == s.npos)
        { // 有限小数
            n = stoull(s.substr(dot + 1));
            m = u64(pow(10, s.length() - dot - 1));
        }
        else
        { // 循环小数
            s.erase(rec, 1);
            n = stoull(s.substr(dot + 1));
            if(dot + 1 != rec)n -= stoull(s.substr(dot + 1, rec - dot - 1));
            m = u64(pow(10, s.length() - rec - 1)) - 1;
            m *= u64(pow(10, rec - dot - 1));
        }
        auto g = gcd(n, m); n /= g; m /= g;
        cout << n + m * a << "/" << m << "\n";
    }

}

typedef long long ll;
using namespace std;
string s;
void solve()
{
        if(s.find("[")==string::npos){
                ll n=s.find(".");
                ll len=s.size()-n-1;
                s=s.erase(s.find("."),1);
                ll f=0;ll g=1;
                for(ll i=0;i<s.size();i++) f=f*10+(s[i]-'0');
                for(ll i=1;i<=len;i++) g*=10;
                ll k=gcd(f,g);
                cout<<f/k<<"/"<<g/k<<endl;
        }
        else{
                ll pos1=s.find("["),pos2=s.find("]");
                ll n=s.find(".");
                if(n==pos1-1){
                        ll f=0,g=0,zheng=0;
                        string str=s.substr(pos1+1,pos2-pos1-1);
                        for(ll i=0;i<str.size();i++){
                                f=f*10+(str[i]-'0');
                        }
                        for(ll i=0;i<str.size();i++){
                                g=g*10+9;
                        }
                        ll k=gcd(f,g);
                        f=f/k,g=g/k;
                        for(ll i=0;i<n;i++){
                                zheng=zheng*10+s[i]-'0';
                        }
                        f=f+zheng*g;
                        k=gcd(f,g);
                        cout<<f/k<<"/"<<g/k<<endl;
                }
                else{
                        string str=s.substr(pos1+1,pos2-pos1-1);
                        ll f=0,g=0,zheng=0;
                        string ssd=s.substr(n+1,pos1-n-1);
                        string spw=ssd+str;
                        string spp=s.substr(0,n);
                        for(ll i=0;i<spp.size();i++) zheng=zheng*10+spp[i]-'0';
                        ll q=0,p=0;
                        for(ll i=0;i<spw.size();i++) q=q*10+spw[i]-'0';
                        for(ll i=0;i<ssd.size();i++) p=p*10+ssd[i]-'0';
                        f=q-p;
                        for(ll i=0;i<str.size();i++) g=g*10+9;
                        for(ll i=0;i<ssd.size();i++) g*=10;
                        ll k=gcd(f,g);
                        f=f/k,g=g/k;
                        f=f+g*zheng;
                        k=gcd(f,g);
                        cout<<f/k<<"/"<<g/k<<endl;
                }
        }
}
int main()
{
        int t;cin>>t;
        for(int j=0;j<t;j++){
                cin>>s;
                cout<<"case #"<<j<<":\n";
                solve();
        }
}

#include<bits/stdc++.h>
using namespace std;
long long int gcd(long long int a,long long int b)
{
    if(a%b==0) return b;
    else return gcd(b,a%b);
}
void solve()
{
  char s[18];
  scanf("%s",s);
  int l=strlen(s);
  double num=0;
  int d=1;
  int flag=1;
  int qi=0;
  int jie=l;
  int chu=0;
  for(int i=0;i<l;i++)
  {
      if(isdigit(s[i])==1&&flag==1) num=num*10+s[i]-'0';
      else if(isdigit(s[i])==1&&flag==0) num=num+(s[i]-'0')/pow(10,d),d++;
      else if(s[i]=='.') flag=0,qi=i+1;
      else if(s[i]=='[') qi=i+1,chu=1;
      else if(s[i]==']') jie=i;
      //cout<<i<<ends<<d<<ends<<num<<endl;
  }
  int k=jie-qi;
  double num2=num*pow(10,k);
  //cout<<d<<endl;
  if(chu==0)
  {
      double k1=pow(10,k);

    long long int num3=num2;
    long long int k2=k1;
    long long int g=gcd(num3,k2);
    //cout<<num3<<ends<<k2<<endl;
    cout<<num3/g<<"/"<<k2/g<<endl;
  }
  else
    {
     double k1=pow(10,k)-1;

     int wei=d-k;
     for(int j=qi;j<jie;j++)
     {
         num2+=(s[j]-'0')/pow(10,wei++);

     }
     num2-=num;
     num2*=pow(10,d-k);
     k1*=pow(10,d-k);
    // cout<<d<<ends<<k<<endl;
    long long int num3=num2;
    long long int k2=k1;
    long long int g=gcd(num3,k2);
    cout<<num3/g<<"/"<<k2/g<<endl;


    }

}
int main()
{
    int n;
    cin>>n;
    for(int i=0;i<n;i++)
    {
        printf("case #%d:\n",i);
        solve();
    }
}

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

LL gcd(LL a, LL b)
{
    if(a > b)
    {
        LL c = a % b;
        while(c)
        {
            a = b;
            b = c;
            c = a % b;
        }
        return b;
    }
    else
        return gcd(b, a);
}

void solve()
{
    string s;
    cin >> s;
    int jdg = 0;
    for(int i = 0; i < s.size(); i++)
    {
        if(s[i] == '[')
        {
            jdg = 1;
            break;
        }
    }
    LL up = 0, down = 0 , up1 = 0, down1 = 0;//up1与down1为循环小数准备

    int pos;//记录小数点的位置
    int pos1, pos2;//记录【】分别的位置

    if(jdg == 0)//有限小数
    {
        for(int i = 0; i < s.size(); i++)
        {
            if(s[i] == '.')
            {
                pos = i;
                break;
            }
        }
        for(int i = 0; i < s.size(); i++)
        {
            if(i == pos)
                continue;
            else
                up = up * 10 + s[i] - '0';
        }
        down = pow (10, s.size() - 1 - pos);
        cout << up / gcd(up, down) << '/' << down / gcd(up, down) << endl;
    }
    else//循环小数
    {
        for(int i = 0; i < s.size(); i++)
        {
            if(s[i] == '.')
                pos = i;
            if(s[i] == '[')
                pos1 = i;
            if(s[i]==']')
                pos2 = i;
        }
        for(int i = 0; i < pos1; i++)
        {
            if(i == pos)
                continue;
            else
                up = up * 10 + s[i] - '0';
        }
        for(int i = pos1 + 1; i < pos2; i++)
        {
            up1 = up1 * 10 + s[i] - '0';
        }
        down1 = pow (10, pos2 - pos1 - 1) - 1;
        if(pos == pos1 - 1)
        {
            up1 = up1 + up * down1;
            cout << up1 / gcd(up1, down1)<< '/' << down1 / gcd(up1, down1) << endl;
        }
        else
        {
            up1 = up1 + up * down1;
            down1 = down1 * pow(10, pos1 - pos - 1);
            cout << up1 / gcd(up1, down1) << '/' << down1 / gcd(up1, down1) << endl;
        }
    }
}
int main()
{
    int t;
    cin >> t;
    for(int i = 0; i < t; i++)
    {
        cout << "case #" << i << ":\n";
        solve();
    }
    return 0;
}

#include <cstddef>
#include <iostream>
#include <istream>
#include <cmath>
#include <string>
using namespace std;

struct rational{
    string origindata;
    string loop; //循环节
    string inter;//整数部分
    string decimal;//小数部分
    long int numer;
    long int denom;
    bool isLoop;
    friend istream& operator>>(istream& is,rational& rat);
    friend ostream& operator<<(ostream& os,const rational& rat);
    void convert();
};
long gcd(long a,long b);
int main(){
    rational rat;
    int T;
    cin >> T;
    for(int i = 0;i < T;i++){
        cin >> rat;
        cout << "case #"<<i<<":\n"<<rat<<endl;
    }
    return 0;
}

istream& operator>>(istream& is,rational& rat){
    is >> rat.origindata;
    size_t dotpos = rat.origindata.find('.');
    size_t looppos = rat.origindata.find('[');
    //整数部分
    if(dotpos == string::npos){
        rat.inter = rat.origindata;
    }else{
        rat.inter = rat.origindata.substr(0,dotpos);
    }

    //小数部分
    if(looppos == string::npos){
        rat.isLoop = false;
        if(dotpos != string::npos)
            rat.decimal = rat.origindata.substr(dotpos+1);
    }else{
        rat.isLoop = true;
        size_t endloop = rat.origindata.find(']');
        rat.decimal = rat.origindata.substr(dotpos+1,looppos - dotpos - 1);
        rat.loop = rat.origindata.substr(looppos+1,endloop - looppos - 1);
    }
    rat.convert();
    return is;
}
ostream& operator<<(ostream& os,const rational& rat){
    if(rat.denom != 1)
        os << rat.numer <<'/'<< rat.denom;
    else
        os << rat.numer;
    return os;
}
void rational::convert(){
    // cout << inter <<' ' << decimal << ' ' << loop << endl;
    long a = pow(10,decimal.length());
    long  b = pow(10,loop.length());
    long small = atol(inter.data())*a + atol(decimal.data());
    long big   = isLoop?small*b+atol(loop.data()):small * 10;
    numer = big - small;
    denom = isLoop?b*a - a:a*9;
    long g = gcd(numer,denom);
    numer /= g;
    denom /= g;
}

long gcd(long a,long b){
    return b ? gcd(b,a%b):a;
}