program p1092;

VAR

i,j,n,m,k,s,p:longint;

c:array[0..100]of 0..1;

f:array[1..100]of int64;

begin

readln(n,m);fillchar(c,sizeof(c),1);

f[1]:=1; c[0]:=0;

for i:=2 to n do f[i]:=f*i;

p:=n;

repeat

dec(p);

k:=m div f[p];

m:=m mod f[p];

i:=0;s:=0;

while s0 then inc(i);

while c[i]=0 do inc(i);

c[i]:=0;

write(i,' ');

until m=0;

for i:=n downto 1 do

if c[i]=1 then write(i,' ');

writeln;

end.

能不能把康拓排序倒过来写！？

额 还以为是全排列呢.....

program p1092;

var

m,s,mm:double;

n,i,j,t,k:longint;

a:array[1..30] of integer;

f:array[1..30] of boolean;

begin

read(n,m);

m:=trunc(m-1);

for i:=n-1 downto 1 do

begin

mm:=m; s:=1;

for j:=i downto 2 do begin s:=s*j; mm:=trunc(mm) div j; end;

a[i]:=trunc(mm);

m:=trunc(m-trunc(mm)*s);

end;

fillchar(f,sizeof(f),true);

for i:=n-1 downto 1 do

begin

t:=0; k:=0;

while (t

为什么会超时啊？？

Var

a,b,p:array[0..20]of integer;

i,j,n,m:longint;

Begin

readln(n,m);

for i:=0 to n-1 do b[i]:=0;

for j:=1 to m-1 do

begin

b[0]:=b[0]+1;

i:=0;

while b[i]>i do

begin

b[i]:=0;

b:=b+1;

i:=i+1;

end;

end;

for i:=0 to n-1 do a[i]:=i+1;

for i:=n-1 downto 0 do

begin

p[i]:=a[b[i]];

for j:=b[i] to i-1 do a[j]:=a[j+1];

end;

for i:=n-1 downto 0 do

begin

write(p[i]);

if i0 then write(' ');

end;

readln;

End.

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

├ 测试数据 06：答案正确... 0ms

├ 测试数据 07：答案正确... 0ms

├ 测试数据 08：答案正确... 0ms

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

Accepted 有效得分：100 有效耗时：0ms

#include

using namespace std;

__int64 d[30];

int f[30];

int t1,n,j,i;

__int64 k;

int main(){

cin>>n>>k;

for(i=1;i=1;i--){d[i]=d*(n-i+1);}

for(i=1;i

var

a:array[1..21]of int64;

b:array[1..20]of integer;

i,j,n:longint;

m,k:int64;

begin

readln(n,m);

m:=m-1;

a[1]:=1;

b[1]:=1;

for i:= 2 to n do

　　 begin

　　 a[i]:=a*i;

　　 b[i]:=i;

　　 end;

for i:=n downto 2 do

　　 begin

　　 k:=m div a;

　　 m:=m mod a;

　　 write(b[k+1],' ');

　　 for j:=k+1 to n-1 do

　　　　b[j]:=b[j+1];

　　 end;

　　 writeln(b[1]);

end.

开一个阶乘数组 然后构造

program p_1;

var

a:array[1..21]of int64;

b:array[1..20]of integer;

i,j,n:longint;

m,k:int64;

begin

readln(n,m);

m:=m-1;

a[1]:=1;

b[1]:=1;

for i:= 2 to n do

begin

a[i]:=a*i;

b[i]:=i;

end;

for i:=n downto 2 do

begin

k:=m div a;

m:=m mod a;

write(b[k+1],' ');

for j:=k+1 to n-1 do

b[j]:=b[j+1];

end;

writeln(b[1]);

end.

傻傻地用全排列算法慢慢找，超时.......

原来是数论的题目.......

超短代码....晒一下

var

a:array[1..21]of int64;

b:array[1..20]of integer;

i,j,n:longint;

m,k:int64;

begin

readln(n,m);

m:=m-1;

a[1]:=1;

b[1]:=1;

for i:= 2 to n do

begin

a[i]:=a*i;

b[i]:=i;

end;

for i:=n downto 2 do

begin

k:=m div a;

m:=m mod a;

write(b[k+1],' ');

for j:=k+1 to n-1 do

b[j]:=b[j+1];

end;

writeln(b[1]);

end.

C似乎装不了20!那么大的数……不过似乎没有那么大的数据？

设一个数组mark[i]记录i有没有被用过

从第一位开始循环 i是第几位

k=(m-1)/jc(n-i)+1 从mark中从前往后找到第k个没用过的数 输出 他就是第i位上的数

m=m%jc(n-i)

如果发现m=0就说明余下的n-i个数都是倒序排列的 直接输出 结束 就可以了

var

n,i,j,k,t:longint;

m:int64;

a:array[1..20] of longint;

s:set of 1..20;

function calc(x:longint):int64;

var i:longint;

ans:int64;

begin

ans:=1;

for i:=1 to x do ans:=ans*i;

exit(ans);

end;

Begin

readln(n,m);

s:=[];

for i:=1 to n do s:=s+[i];

m:=m-1;

for i:=1 to n do

begin

for j:=n downto 1 do

if j in s then begin

t:=0;

for k:=1 to j-1 do if k in s then inc(t);

if t*calc(n-i)>m then continue;

s:=s-[j];

a[i]:=j;

m:=m-t*calc(n-i);

break;

end;

end;

for i:=1 to n do write(a[i],' ');

end.

var

n :longint;

fac :array[0..20]of int64;

v :array[0..20]of boolean;

m :int64;

k,s :longint;

i,j :longint;

begin

readln(n,m);

fac[0]:=1;

for i:=1 to n do

fac[i]:=fac*i;

fillchar(v,sizeof(v),true);

for i:=n downto 1 do begin

j:=1;

while j*fac

作个康托展开

const jc:array[1..20] of int64=

(

1,

2,

6,

24,

120,

720,

5040,

40320,

362880,

3628800,

39916800,

479001600,

6227020800,

87188291200,

1307674368000,

20922789888000,

355687428096000,

6402373705728000,

121645100408832000,

2432902008176640000

);

// jc[] 为阶乘表

var a:array[1..20] of integer;

n,b,c:longint;

p,m:int64;

list:array[1..20] of boolean;

function l(num:integer):integer;

var p:integer;

begin

for p:=1 to n do

begin

if list[p]=false then dec(num);

if num=0 then break;

end;

l:=p;

list[l]:=true;

end;

begin

fillchar(list,sizeof(list),false);

readln(n,m);

p:=1;

m:=m-1;

while p

哎，想不A都不行，什么世道。。。。。。

var

used:array [1..20] of boolean;

a:array [1..20] of integer;

m:int64;

n:longint;

s,st:string;

procedure solve(dep:integer);

var

flag:boolean;

i,j,k:longint;

t:int64;

begin

flag:=true;

if dep>n then

begin

for i:=1 to (n-1) do

write(a[i],' ');

writeln(a[n]);

flag:=false;

end;

if flag then

begin

t:=1;

for i:=2 to (n-dep) do

t:=t*i;

if m mod t=0 then

j:=m div t

else

j:=m div t+1;

k:=0;

for i:=1 to n do

if not used[i] then

begin

inc(k);

if k=j then

begin

a[dep]:=i;

used[i]:=true;

break

end;

end;

m:=m mod t;

if m=0 then

m:=t;

solve(dep+1);

end;

end;

begin

readln(st);

s:=copy(st,1,pos(' ',st)-1);

st:=copy(st,pos(' ',st)+1,length(st));

val(s,n);

val(st,m);

fillchar(used,sizeof(used),false);

solve(1);

end.

前面的大牛似乎程序都很长。。。。。。。

来一个短的^_^

（一次AC。。。）

======================晒程序=====================

var

d:array[0..30] of int64;

f:array[0..30] of integer;

t1,n,j,i:LONGINT;

k:int64;

begin

readln(n,k);

for i:=1 to n do f[i]:=i;

d[n]:=1;d[n+1]:=1;

for i:=n-1 downto 1 do d[i]:=d*(n-i+1);

for i:=1 to n do begin

t1:=((k-1) div d)+1;

k:=((k-1)mod d)+1;

write(f[t1],' ');

for j:=t1 to n-i+1 do f[j]:=f[j+1];

end;

writeln();

end.

Accepted 有效得分：100 有效耗时：0ms

大概O(N)吧。。。。

^_^

终于真正地能一次AC了……

用构造解，基本做出来就不会超时了。还有考虑边界。

program dsa;

var a:array[1..20]of longint;

i,j:longint;

bb,n,k:int64;

m:qword;

begin

readln(n,m);

bb:=1;

for i:=2 to n-1 do bb:=bb*i;

for i:=1 to n do a[i]:=i;

i:=1;

if m=1 then

begin

for j:=1 to n do write(a[j],' ');

exit;

end;

while i0 do

begin

m:=m-bb;

inc(k);

end;

write(a[k],' ');

for j:=k to n-1 do a[j]:=a[j+1];

bb:=bb div (n-i);

inc(i);

end;

end.

农夫山泉