不能输出x/1 ,出题目不用这样吧，太不讲道理了。而且题目中说是连分数和分数的互化，这样写本质上没错啊！！！

小鸟直接模拟居然写了这么长的程序。。。

program p1;

var

ch:char;

procedure swap(var a,b:longint);

var

k:longint;

begin

k:=a;

a:=b;

b:=k;

end;

procedure chuli1;

var

a:array[1..200]of longint;

z,m,i,j,k:longint;

begin

fillchar(a,sizeof(a),0);

read(ch);

i:=0;

while ch']' do

begin

inc(i);

while (ch',' )and ( ch';') and (ch']') and (ch'[') do

begin a[i]:=a[i]*10+(ord(ch)-ord('0')); read(ch); end;

if ch=']' then break;

read(ch);

end;

readln;

m:=a[i]; z:=1;

while i>1 do

begin

dec(i);

z:=z +m *a[i];

swap(m,z);

end;

swap(m,z);

if m>z then j:=trunc(sqrt(m)) else j:=trunc(sqrt(z));

for k:=2 to j do

if (m mod k =0) and(z mod k =0) then

while (m mod k =0) and(z mod k =0) do

begin

m:=m div k;

z:=z div k;

end;

if m1 then

writeln(z,'/',m) else writeln(z);

end;

procedure chuli2;

var

s,z,m,i,j,k:longint;

begin

z:=0;

while ch'/' do

begin

z:=z*10+(ord(ch)-ord('0'));

read(ch);

end;

read(m);

readln;

s:=z div m;

write('[',s,';');

z:=z-m*s;

swap(m,z);

if(m mod z0) then

begin

s:=z div m;

write(s);

z:=z-m*s;

swap(m,z);

end;

while (m 0) do

begin

s:=z div m;

write(',',s);

z:=z-m*s;

swap(m,z);

end;

writeln(']');

end;

begin

while not eof do

begin

read(ch);

if ch='[' then chuli1 else chuli2;

end;

end.

var

s:string;

function gd(m,n:longint):longint;

begin

if n=0 then

gd:=m

else

gd:=gd(n,m mod n);

end;

procedure a;

var

i,tot,fz,fm,tmp,z,code,k:longint;

x:array[1..1000] of longint;

begin

k:=pos(';',s);

if k=0 then

begin

writeln(copy(s,2,length(s)-2));

exit;

end;

if s[k+1]=']' then

begin

writeln(copy(s,2,k-2));

exit;

end;

val(copy(s,2,k-2),z,code);

delete(s,1,k);

delete(s,length(s),1);

k:=1;

tot:=0;

while true do

begin

inc(tot);

k:=pos(',',s);

if k0 then

begin

val(copy(s,1,k-1),tmp,code);

x[tot]:=tmp;

delete(s,1,k);

end

else

begin

val(s,tmp,code);

x[tot]:=tmp;

break;

end;

end;

i:=tot;

fm:=x[tot];

fz:=1;

if i1 then

repeat

fz:=fm*x+fz;

tmp:=fz;

fz:=fm;

fm:=tmp;

dec(i);

until i=1;

fz:=fz+fm*z;

tmp:=gd(fz,fm);

fz:=fz div tmp;

fm:=fm div tmp;

if fm=1 then

writeln(fz)

else

writeln(fz,'/',fm);

end;

procedure b;

var

i,tot,fz,fm,tmp,z,zz,code,k:longint;

x:array[1..1000] of longint;

begin

k:=pos('/',s);

if k=0 then

begin

writeln(s);

exit;

end;

val(copy(s,1,k-1),fz,code);

val(copy(s,k+1,length(s)-k),fm,code);

tmp:=gd(fz,fm);

fz:=fz div tmp;

fm:=fm div tmp;

z:=0;

while fz>=fm do

begin

fz:=fz-fm;

inc(z);

end;

if fz=0 then

begin

write('[',z,']');

exit;

end;

write('[',z,';');

zz:=0;

repeat

tmp:=fm;

fm:=fz;

fz:=tmp;

zz:=0;

while fz>fm do

begin

fz:=fz-fm;

inc(zz);

end;

write(zz,',');

until fz=1;

write(fm,']');

writeln;

end;

begin

while not eof do

begin

readln(s);

if s[1]='[' then

a

else

b;

end;

end.

测一下[3;3]就知道了……

在昨天的程序里

加一个补丁

就行了

hint：昨天80分……

编译通过...

├ 测试数据 01：答案正确... 0ms

├ 测试数据 02：答案正确... 0ms

├ 测试数据 03：答案正确... 0ms

├ 测试数据 04：答案正确... 0ms

├ 测试数据 05：答案正确... 0ms

---|---|---|---|---|---|---|---|-

Accepted 有效得分：100 有效耗时：0ms

var

s:string; b:boolean;

i,j,k,n,i1,i2,x,y,z1,z2:longint;

a:array[0..10000]of longint;

function gcd(x,y:longint):longint;

begin

if y=0 then exit(x);

gcd:=gcd(y,x mod y);

end;

begin

while not(eof)do

begin

readln(s); b:=false;

if pos('[',s)0 then

if pos(';',s)=0 then writeln(copy(s,2,length(s)-2))

else

begin fillchar(a,sizeof(a),0);

i1:=pos(';',s);

if s[2]='-' then begin b:=true;val(copy(s,3,i1-3),a[0]);end

else val(copy(s,2,i1-2),a[0]);

delete(s,1,pos(';',s));i2:=0;

while pos(',',s)0 do

begin

inc(i2);

val(copy(s,1,pos(',',s)-1),a[i2]);

delete(s,1,pos(',',s));

end;

inc(i2);val(copy(s,1,length(s)-1),a[i2]);

x:=1;y:=a[i2];

for j:=i2-1 downto 0 do

begin

z1:=a[j]*y+x;

x:=y;

y:=z1;

end;

if b then writeln(y-2*x*a[0],'/',x)else

writeln(y,'/',x);

end

else if pos('/',s)0then

if pos('/1',s)=length(s)-1 then writeln('[',copy(s,1,length(s)-2),']')

else begin

i1:=pos('/',s); if pos('-',s)0then begin val(copy(s,2,i1-2),x); b:=true;end

else begin val(copy(s,1,i1-1),x);end;

val(copy(s,i1+1,length(s)-i1),y);

z1:=gcd(x,y); write('[');

x:=x div z1;y:=y div z1;

if b then begin

write('-',x div y+1,';');

x:=y*(x div y+1)-x;end else

write(x div y,';'); x:=x mod y;

while (y0)and(y1) do

begin

if (y mod x0)then

write(y div x,',')else write(y div x);

y:=y mod x;

z2:=y;y:=x;x:=z2;

end;

writeln(']');

end;

end;

end.

话说这个题还真让我迷茫呢

program test1;

var s:string;

t1,t,t2,p,code,ls,up,down:longint;

begin

while not eof do

begin

readln(s);

if s[1]='[' then begin

ls:=length(s);

dec(ls);

p:=ls;

while s[p] in ['0'..'9'] do dec(p);

val(copy(s,p+1,ls-p),t1,code);

up:=1;down:=t1;

ls:=p;

if pos(';',s)=0 then begin writeln(t1);continue;end;

repeat

p:=ls-1;

while s[p] in ['0'..'9'] do dec(p);

val(copy(s,p+1,ls-p-1),t1,code);

up:=up+t1*down;

t:=up;up:=down;down:=t;

ls:=p;

until ls=1;

t:=up;up:=down;down:=t;

writeln(up,'/',down);

end

else begin

p:=pos('/',s);

val(copy(s,1,p-1),t1,code);

delete(s,1,p);

val(s,t2,code);

write('[');

write(t1 div t2);

if t1 mod t2=0 then begin writeln(']');continue;end

else write(';');

repeat

t1:=t1 mod t2;

t:=t1;t1:=t2;t2:=t;

write(t1 div t2);

if t1 mod t20 then write(',');

until t1 mod t2=0;

writeln(']');

end;

end;

end.

额...

沙茶了...

果然 不能输出 x/1 这种东西 不讲道理呀....

program no1;

var st:string;

function zzxc(m,n:int64):int64;

var a,b,r:int64;

begin

a:=m;b:=n;

repeat

r:=a mod b;

a:=b; b:=r;

until r=0;

zzxc:=a;

end;

procedure doit(st:string);

var s,s1,s2,ss1,ss2:int64;

j,k,i:longint;

a:array[0..200] of longint;

bz:boolean;

begin

if st[1]='[' then

begin

fillchar(a,sizeof(a),0);

j:=2;k:=0;

while j='0') and (st[j]

我比赛的时候考虑到分子很大，分母很小的情况。于是计算不大于某个数的整数时二分了。结果对于某些分母较大的整数，在二分过程中会超过longint。改成循环后80分，结果发现漏了一种情况，即[x]是不用数出分母1的（考虑了约分竟然没考虑这个）。

program test1;

const re:set of char=(['0','1','2','3','4','5','6','7','8','9']);

var i,j,k,m:longint;

s1,s2:ansistring;

function gcd(a,b:longint):longint;

begin

if b=0 then gcd:=a else gcd:=gcd(b,a mod b);

end;

procedure work1;

var i,j,k,m,n,fz,fm,t:longint;

x:array[1..10000] of longint;

boo:boolean;

Begin

i:=2;j:=1; boo:=false;

if s1[i]='-' then begin inc(i);boo:=true;end;

while ifm then

begin

for i:= n-1 downto 1 do

begin

t:=fm;fm:=fz;fz:=t;

fz:=fz+x[i]*fm;

end;

m:=gcd(fm,fz);

fm:=fm div m;

fz:=fz div m;

end;

if fm

完美版

function gcd(p,q:longint):longint;

var

temp:longint;

begin

if q=0 then gcd:=p

else gcd:=(q,p mod q);

end;

begin

while not eof do

begin

readln(s);

len:=length(s);

fillchar(a,sizeof(a),0);

if s[1]='[' then

begin

num:=1;ex:=10;

for i:=2 to len-1 do

begin

if (s[i]=';') or (s[i]=',') then inc(num)

else

begin

a[num]:=a[num]*ex+ord(s[i]-'0');

end;

end;

if num=1 then writeln(a[1])

else

begin

up:=1;down:=a[num];

for i:=num-1 downto 1 do

begin

up:=up+a[i]*down;

r:=up;up:=down;down:=r;

g:=gcd(up,down);

up:=up div g;down:=down div g;

end;

if up1 then

writeln(down,'/',up)

else writeln(down);

end;

end

else

begin

l:=0;

while s[l]'/' do inc(l);

ex:=10;pp:=0;dd:=0;

for i:=1 to l-1 do

pp:=pp*ex+ord(s[i]-'0');

ex:=10;

for i:=l+1 to len do

dd:=dd*ex+ord(s[i]-'0');

write('[');

g:=gcd(pp,dd);ex:=1;

pp:=pp div g;dd:=dd div g;

while dd1 do

begin

if ex=1 then

begin

write(pp div dd,';');

inc(ex);

end

else write(pp div dd,',');

pp:=pp-pp div dd*dd;

temp:=pp;pp:=dd;dd:=temp;

end;

writeln(pp,']');

end;

end;

end.

2/1到底是什么啊啊啊啊

编译通过...

测试数据01：答案正确... 0ms

测试数据02：答案正确... 0ms

测试数据03：答案正确... 0ms

测试数据04：答案错误...程序输出比正确答案长

测试数据05：答案正确... 0ms

---|---|---|---|---|---|---|---|-

Unaccepted 有效得分：80 有效耗时：0ms

why??????????????????

不知道哪里错了。。和一个AC的程序对拍一直是正确的。。。悲剧。

#include

#include

char buffer[2001],*buf;

inline long upfloor(double x)

{

if(x-(long)x-0.00000000000000001)return (long)x+1;

return (long)x;

}

inline long getw(long x)

{

if(!x)return 1;

if(x>0)

return upfloor(log10(x));

return upfloor(log10(-x))+1;

}

void ctof(long &a,long &b)

{

long t,t1,t2;

if(sscanf(buf,"%d",&t)>0)

{

buf+=getw(t)+1;

ctof(t1,t2);

b=t1;

a=t*t1+t2;

}

else

{

a=1;b=0;

}

}

void ftoc(long a,long b)

{

if(!a)return ;

printf("%d,",b/a);

b%=a;

ftoc(b,a);

}

void inline swap(long &a,long &b)

{

long t=a;

a=b;

b=t;

}

long Euclid(long a,long b)

{

if(ab?a-b:b-a,a>b?b:a);

else if(b&1)return Euclid(a>>1,b);

else if(a&1)return Euclid(a,b>>1);

else return Euclid(a>>1,b>>1);

}

void yf(long& a,long& b)

{

long register tmp=Euclid(a,b);

a/=tmp;b/=tmp;

}

int main()

{

freopen("fraction.in","r",stdin);

freopen("fraction.out","w",stdout);

while(scanf("%s",buffer)!=EOF)

if(buffer[0]=='[')

{

long a,b,t;

sscanf(buffer,"[%d",&t);

buf=buffer+getw(t)+2;

ctof(a,b);

yf(a,b);

if(a-1)

printf("%d/%d\n",b+t*a,a);

else

printf("%d\n",b+t);

}

else

{

long a,b;

sscanf(buffer,"%d/%d\n",&a,&b);

//printf("%d/%d=",a,b);

if(b=0)

{

yf(a,b);

printf("[%d;",a/b);

a%=b;

}

else

{

long tmp=-a;

yf(tmp,b);

if(tmp%b)

{

printf("[-%d;",tmp/b+1);

a=b-tmp%b;

}

else

{

printf("[-%d;",tmp/b);

a=0;

}

}

ftoc(a,b);

fseek(stdout,-1,SEEK_CUR);

printf("]\n");

}

return 0;

}

编译通过...

├ 测试数据 01：答案正确... 0ms

├ 测试数据 02：答案正确... 0ms

├ 测试数据 03：答案正确... 0ms

├ 测试数据 04：答案正确... 0ms

├ 测试数据 05：答案正确... 0ms

---|---|---|---|---|---|---|---|-

Accepted 有效得分：100 有效耗时：0ms

后来！

编译通过...

├ 测试数据 01：答案正确... 0ms

├ 测试数据 02：答案正确... 0ms

├ 测试数据 03：答案正确... 0ms

├ 测试数据 04：答案错误...程序输出比正确答案长

├ 测试数据 05：答案正确... 0ms

---|---|---|---|---|---|---|---|-

Unaccepted 有效得分：80 有效耗时：0ms

为啥？

调了半天，原来是分母为1的时候不输出

交了n多次

占楼~~orz

*|可能不是满分程序

*|也可能是

#include

#include

char s[10000];

int main()

{

for(;scanf("%s",s)!=-1;)

{

if(s[0]=='[')//连分数---|>分数

{

int j=0,locate=1,number;

long turn[10000]={};

for(number=0;locate='0'&&s[locate]=0;j--)

{

p1=p1+turn[j]*p2;

int temp=p1;

p1=p2;

p2=temp;

}

int temp=p1;

p1=p2;

p2=temp;

for(j=p2;j>0;j--)

if(p1%j==0&&p2%j==0)

{

p1/=j;

p2/=j;

break;

}

if(p2>1)printf("%d/%d\n",p1,p2);

else printf("%d\n",p1);

}

else

{

int j=0,locate=0,number;

long turn[2]={};

for(number=0;locate='0'&&s[locate]

mao2

我算什么？