var a:array[0..10000] of longint;
    i,n,m,j,p:longint;
    ans:qword;
    
    procedure swap(var a,b:longint);
    var t:longint;
    begin
        t:=a;a:=b;b:=t;
    end;
    
    procedure kuai(l,r:longint);
var i,j,x,y:longint;
begin
i:=l;j:=r;
x:=a[(l+r) div 2];
repeat
while a[i]<x do inc(i);
while a[j]>x do dec(j);
if i<=j then
begin
y:=a[i];a[i]:=a[j];a[j]:=y;inc(i);dec(j);
end;
until i>j;
if i<r then kuai(i,r);
if l<j then kuai(l,j);
end;
    
begin
    {assign(input,'martian.in');assign(output,'martian.out');
    reset(input);rewrite(output);}
    readln(n,m);
    for i:=1 to n do read(a[i]);
    for m:=1 to m do begin
        for j:=n downto 1 do if a[j]>=a[pred(j)] then break;
        for p:=n downto 1 do if a[pred(j)]<=a[p] then break;
        swap(a[pred(j)],a[p]);
        kuai(j,n);
    end;
    for i:=1 to n do write(a[i],' ');
    //close(input);close(output);
end.

/*刚开始看还没理解题目意思，但是仔细一想才发现，其实 m的真正含义是讲n个数全排列第 m+1小的数 
例如，m=1，样例中，那么输出的是第2小的，也可以说是除去本身之后全排列 第m小的排列组合*/
#include<cstdio>
#include<algorithm>
#include<iostream>
int n,m,num[10001];
int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;++i)scanf("%d",&num[i]);
    for(int i=1;i<=m;++i)
    {
    //详见于 “代码大全 -中 -算法- next_permutation函数 ” 的代码 
    
/***********************************************************************************有可能看不懂 next_permutation函数
    附上 next_permutation函数的代码 正确理解 next_permutation函数 可见于网址：http://blog.sina.com.cn/s/blog_9f7ea4390101101u.html 
    #include<iostream>
    #include<cstdio>
    #include<algorithm>
    using namespace std;
    int main()
    {
    int n;
 int a[n];
 scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
 do
{
    for(int i=0;i<n;i++)
     cout<<a[i]<<" ";
     cout<<endl;
} while (next_permutation(a,a+5));//表示对前5个数进行排列，后面的数不管 
//next_permutation(a,a+3)  为对前面前三个数排列组合 
return 0;
}
    *******************************************************************************************************************/
    std::next_permutation(num+1,num+n+1);//第m次排列出来的 
    /*for(int i=1;i<=n;i++)printf("%d ",num[i]);
    printf("\n");*/
    }
//    printf("\n");
    for(int i=1;i<=n;++i)printf("%d ",num[i]);
    return 0;
}

看你们都用STL大法，看来我只有手写生成字典序排列了。。。
编译成功

foo.c: In function 'main':
foo.c:7:27: warning: 'min' may be used uninitialized in this function [-Wmaybe-uninitialized]
int n, m, i, j, temp, t, min;
^

测试数据 #0: Accepted, time = 0 ms, mem = 524 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 528 KiB, score = 10

测试数据 #2: Accepted, time = 0 ms, mem = 524 KiB, score = 10

测试数据 #3: Accepted, time = 0 ms, mem = 524 KiB, score = 10

测试数据 #4: Accepted, time = 0 ms, mem = 528 KiB, score = 10

测试数据 #5: Accepted, time = 0 ms, mem = 528 KiB, score = 10

测试数据 #6: Accepted, time = 0 ms, mem = 524 KiB, score = 10

测试数据 #7: Accepted, time = 0 ms, mem = 524 KiB, score = 10

测试数据 #8: Accepted, time = 0 ms, mem = 528 KiB, score = 10

测试数据 #9: Accepted, time = 0 ms, mem = 528 KiB, score = 10

Accepted, time = 0 ms, mem = 528 KiB, score = 100

代码

include <stdio.h>

define SWAP(a,b,t) t=a, a=b, b=t

int a[10005];

int main (void)
{
int n, m, i, j, temp, t, min;
scanf("%d%d",&n,&m);
for (i=0; i<n; i++) {
scanf("%d",a+i);
}
while (m--) {
for (i=n-1; i>0; i--) {
if (a[i] > a[i-1]) {
for (j=n-1; j>=i; j--) {
if (a[j] > a[i-1]) {
min = j;
break;
}
}
SWAP(a[i-1], a[min], temp);
if (i < n-1) {
t = (n-i)/2 + i;
for (j=i; j<t; j++) {
SWAP(a[j], a[n+i-1-j], temp);
}
}
break;
}
}
}
for (i=0; i<n; i++) {
printf(i==0 ? "%d" : " %d",a[i]);
}
putchar('\n');
return 0;
}

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
#include <algorithm>
#include <math.h>

using namespace std;
int a[10005];
int n,m;
int main(){
scanf("%d%d",&n,&m);
for(int i = 1;i <= n;i ++){
scanf("%d",&a[i]);
}
for(int i = 1;i <= m;i ++){
next_permutation(a + 1,a + n + 1);
}
for(int i = 1;i <= n;i ++){
printf("%d ",a[i]);
}
return 0;
}
STL666

#include<cstdio>
#include<algorithm>
int n,m,num[10001];
int main(){
scanf("%d%d",&n,&m);
for(int i=1;i<=n;++i)scanf("%d",&num[i]);
for(int i=1;i<=m;++i)std::next_permutation(num+1,num+n+1);
for(int i=1;i<=n;++i)printf("%d ",num[i]);
}
STL大法好！！！

program p1115;
var n,m,i,j,k:longint;
a,b:array[0..10000] of longint;
procedure work;
var i,t,p:longint;
begin
for i:=n downto 1 do
if a[i-1]<a[i] then break;
t:=i-1;
for i:=n downto t+1 do
if a[i]>a[t] then break;
p:=a[i];a[i]:=a[t];a[t]:=p;
for i:=t+1 to n do b[i]:=a[n+t+1-i];
for i:=t+1 to n do a[i]:=b[i];
end;
//main
begin
readln(n);
readln(m);
for i:=1 to n do read(a[i]);
for k:=1 to m do work;
for i:=1 to n-1 do write(a[i],' ');
writeln(a[n]);
end.

std::next_permutation
Rearranges the elements in the range [first,last) into the next lexicographically greater permutation.

该函数定义在algorithm头文件中，可以将序列变换为下一个字典序的排列。

细看此题之前的题解，“柳暗花明又一村”，同时也是时代变化的体现。

/*************************************************************************
> File Name: 1115.cpp
> Author: Netcan
> Blog: http://www.netcan.xyz
> Mail: 1469709759@qq.com
> Created Time: Fri 01 May 2015 21:02:36 CST
************************************************************************/

#include <iostream>
#include <vector>
#include <string>
#include <queue>
#include <algorithm>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <sstream>
#include <deque>
#include <functional>
#include <iterator>
#include <list>
#include <map>
#include <memory>
#include <stack>
#include <set>
#include <numeric>
#include <utility>
#include <cstring>
using namespace std;

int main()
{
int N,M;
int a[10005];
while(scanf("%d%d", &N, &M)==2) {
for(int i=0; i<N; ++i)
scanf("%d", &a[i]);
for(int i=0; i<M; ++i)
next_permutation(a, a+N);
for(int i=0; i<N; ++i)
printf("%d ", a[i]);
printf("\n");

}

return 0;
}

我想多了，千万别用康托法！！

我只知道用模拟算法能够通过。就从1加到100，挨个走。
#include<iostream>
#include<string.h>
#include<math.h>
using namespace std;
int a[10003];
int size;
void sort(int to){
int i, j;
for (i = 0; i < to; i++){
for (j = i + 1; j < to; j++){
if (a[i] < a[j]){
int temp = a[j];
a[j] = a[i];
a[i] = temp;
}
}
}
}
void inc(){
int i, j, k;
for (i = 1; i < size; i++){
if (a[i - 1]>a[i])break;
}
for (j = 0; j < i; j++){
if (a[j]>a[i])break;
}
int temp = a[i];
a[i] = a[j];
a[j] = temp;
sort(i);
}
int main(){
freopen("in.txt", "r", stdin);
int adding;
int i, j, k;
cin >> size >> adding;
for (i = size - 1; i >= 0; i--)cin >> a[i];
for (i = 0; i < adding; i++){
inc();
}
for (j = size - 1; j >= 0; j--)cout << a[j] << " ";
return 0;
}

全是神犇。。。**orzorz**

测试数据 #0: Accepted, time = 15 ms, mem = 856 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 852 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 856 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 852 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 856 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 856 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 852 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 852 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 856 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 856 KiB, score = 10
Accepted, time = 15 ms, mem = 856 KiB, score = 100

program hxr;
var
a:array[1..10005] of longint;
n,m,k:integer;
i,j,p,p1,min:longint;
procedure sort(l,r: longint);
var
i,j,x,y: longint;
begin
i:=l;
j:=r;
x:=a[(l+r) div 2];
repeat
while a[i]<x do
inc(i);
while x<a[j] do
dec(j);
if not(i>j) then
begin
y:=a[i];
a[i]:=a[j];
a[j]:=y;
inc(i);
j:=j-1;
end;
until i>j;
if l<j then
sort(l,j);
if i<r then
sort(i,r);
end;
begin
readln(n);
readln(m);
for i:=1 to n do
read(a[i]);
for i:=1 to m do
begin
for j:=n-1 downto 1 do
if a[j]<a[j+1] then begin
p:=j;
break;
end;
min:=maxint;
for j:=p to n do
if (a[j]>a[p])and(min>a[j]) then begin
min:=a[j];
p1:=j;
end;
k:=a[p];
a[p]:=a[p1];
a[p1]:=k;
sort(p+1,n);
end;
for i:=1 to n do
write(a[i],' ');
end.

var
n,k,i,ans:longint;
a:array[1..10000] of longint;

procedure qsort(l,r:longint);
var
i,j,mid,t:longint;
begin
i:=l;
j:=r;
mid:=a[(l+r) div 2];
repeat
while a[i]<mid do inc(i);
while a[j]>mid do dec(j);
if i<=j then
begin
t:=a[i];
a[i]:=a[j];
a[j]:=t;
inc(i);
dec(j);
end;
until i>j;
if l<j then qsort(l,j);
if i<r then qsort(i,r);
end;

procedure try(p:longint);
var
i,k,l,j,t:longint;
t1:boolean;
begin
if ans=0 then
begin
for i:=1 to n-1 do
write(a[i],' ');
writeln(a[n]);
halt;
end;
for i:=n-1 downto p+1 do
begin
t1:=true;
while t1 do
begin
t1:=false;
l:=maxlongint;
k:=0;
for j:=i+1 to n do
if (a[j]>a[i]) and (a[j]<l) then
begin
l:=a[j];
k:=j;
end;
if k<>0 then
begin
t:=a[i];
a[i]:=a[k];
a[k]:=t;
t1:=true;
qsort(i+1,n);
dec(ans);
try(i);
end;
end;
end;
end;

begin
readln(n);
readln(k);
for i:=1 to n do
read(a[i]);
ans:=k;
try(0);
end.