思路：
无循环小数：直接gcd没啥好说的
纯循环小数:一个循环节有几个数,分母就有几个9,分子则为一个循环节上的数
例.0.3=3/9, 0.347=347/999
混循环小数,循环节有几个数,分母就有几个9,不循环的有几个数,分母再添几个0,分子是从不循环到一个循环节数减去不循环的数
例.0.32=(32-3)/90, 0.2134=(2134-21)/9900
from：https://wenku.baidu.com/view/44e00a25ddccda38366baf0d.html
2980. 小数转化分数
CCXXXI_
stoull是个好东西，可以省去不少手动string转int的麻烦
#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";
}
}
Andrew-Malcom
include
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();
}
}
Master X
循环小数系列终于差不多了……
这种题简直有剧毒……mmp
需要注意的细节很多
附上巨蠢的代码造福后人……
#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();
}
}
Saitama
#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;
}
10175102259
注意：无法AC的同学除了算法的问题还要注意自己是不是用了容量足够大的数据类型来计算，计算结果有没有超出范围，还有如果使用long long int的话（C）记得写成%lld
10195101505
#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;
}
写的很丑
10152130230
写出了，过不掉。不管了，贴代码，供参考
6666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666[em:03]
写的比较长，但是思路还算清楚，这种类型的题只要有耐心把情况细分，不用急于想出一种高明的方法把所有情况都包含在内，都是不难的。