permutation 1【HDU-6628】【排列组合+线段树】

  这次是杭电多校的第五场了，时间也已经到了一半的时候，但是，这场打的很糟糕了。

  这道题我比赛的时候，一直再想办法完善我的思维， 我手玩了几组样例，2、3、4，然后找到了个看似很成立的规律就是我们可以固定前两个数，后面的N-2个数就是按照阶乘的某种关系，我们可以就寻访第K小，然后推下去，想想，no problem！开始！然后WA，一直WA…… QAQ，然后最后一发略改了精度再交，比赛已经结束了。

  下来之后想到这样一组可以推翻我自己的样例：

[code]1
5 25

true answer：5 3 4 1 2

my answer：4 2 1 3 5

  是因为，我根据"(N-2) ! "来确定了位置，但是实际上并不是这样来确定的！

Wrong Code：

[code]#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef long long ll;
const int maxN = 25;
int N;
ll K, jiecheng[maxN], pre_sum[105];
int tree[maxN<<2];
void buildTree(int rt, int l, int r)
{
tree[rt] = 1;
if(l == r) return;
int mid = (l + r)>>1;
buildTree(rt<<1, l, mid);
buildTree(rt<<1|1, mid + 1, r);
tree[rt] = tree[rt<<1] + tree[rt<<1|1];
}
int query(int rt, int l, int r, int kth)
{
if(l == r) return l;
int mid = (l + r)>>1;
if(tree[rt<<1] >= kth) return query(rt<<1, l, mid, kth);
else return query(rt<<1|1, mid + 1, r, kth - tree[rt<<1]);
}
void update(int rt, int l, int r, int qx)
{
tree[rt]--;
if(l == r) return;
int mid = (l + r)>>1;
if(qx <= mid) update(rt<<1, l, mid, qx);
else update(rt<<1|1, mid + 1, r, qx);
}
int main()
{
int T; scanf("%d", &T);
jiecheng[0] = 1;
jiecheng[1] = 1;
for(ll i=2; i<=20; i++) jiecheng[i] = jiecheng[i - 1] * i;
pre_sum[0] = 0;
while(T--)
{
scanf("%d%lld", &N, &K);
buildTree(1, 1, N);
for(int i=1; i<=N-1; i++) pre_sum[i] = pre_sum[i-1] + i;
for(int i=1; i<=N-1; i++) pre_sum[i + N - 1] = pre_sum[i + N - 2] + N - i;
ll Big_pos = K / jiecheng[N - 2] + (K % jiecheng[N - 2] == 0 ? 0 : 1); //大区间
int pos = (int)(lower_bound(pre_sum + 1, pre_sum + 2 * N - 2, Big_pos) - pre_sum);
K -= pre_sum[pos - 1] * jiecheng[N - 2];
int rang_len;
if(pos < N) rang_len = N - pos;
else rang_len = pos - N + 1;
ll small_pos = K / jiecheng[N - 2];
if(small_pos * jiecheng[N-2] == K) small_pos --;
small_pos++;
K -= (small_pos - 1) * jiecheng[N-2];
int a = N - (int)small_pos + 1, b = a - rang_len;
if(pos < N) printf("%d %d", a, b);
else printf("%d %d", b, a);
update(1, 1, N, a);
update(1, 1, N, b);
for(int i=N-3; i>=0; i--)
{
int tmp = K / jiecheng[i] - (K % jiecheng[i] == 0 ? 1 : 0), now;
//tmp = min(tmp, tree[1] - 1);
printf(" %d", (now = query(1, 1, N, tmp + 1)));
update(1, 1, N, now);
if(K > tmp * jiecheng[i]) K -= tmp * jiecheng[i];
}
printf("\n");
}
return 0;
}

  现在，我要做的是，去把这道题改正回来，我们看到这里的K只有最大1e4，也知道8！= 40320就是大于1e4了，所以，我们不妨从这个角度出发来思考这个问题。

loading…… 

  接下去，我发现我的这样的写法只局限于固定前两个数的时候，并且恒定成立那么，我们是不是可以对其余的进行直接暴力排序即可，因为在这里，我们要让"(N - 2) ! ≥ 1e4"，所以，N需要>9，也就是N从10开始的所有的答案都是可以符合我上面的代码的思维的，接下去我们只需要去暴力一个预处理"≤9"的所有的数的这样的第K小的状态即可。

Accept Code:

[code]#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxN = 25;
bool vis[maxN];
int N, K, tot, arr[maxN], pre[10][10005][10], ans[maxN];
struct node
{
int a[maxN], len, d[maxN];
friend bool operator < (node e1, node e2)
{
for(int i=1; i<e1.len; i++)
{
if(e1.d[i] < e2.d[i]) return true;
else if(e1.d[i] > e2.d[i]) return false;
}
return false;
}
}a[362885];
ll jiecheng[maxN];
void dfs(int now, int depth)
{
if(now == depth)
{
++tot;
for(int i=1; i<now; i++) a[tot].d[i] = arr[i+1] - arr[i];
for(int i=1; i<=now; i++) a[tot].a[i] = arr[i];
a[tot].len = depth;
return;
}
for(int i=1; i<=depth; i++)
{
if(!vis[i])
{
vis[i] = true;
arr[now + 1] = i;
dfs(now + 1, depth);
vis[i] = false;
}
}
}
inline void pre_did()
{
jiecheng[0] = jiecheng[1] = 1;
for(int i=2; i<=20; i++) jiecheng[i] = jiecheng[i-1] * i;
for(int i=2; i<=9; i++)
{
memset(vis, false, sizeof(vis));
tot = 0;
dfs(0, i);
sort(a + 1, a + tot + 1);
int _UP = min(10000, tot);
for(int j=1; j<=_UP; j++)   //第几小
{
for(int k=1; k<=i; k++)
{
pre[i][j][k] = a[j].a[k];
}
}
}
}

int tree[maxN<<2];
void buildTree(int rt, int l, int r)
{
tree[rt] = 1;
if(l == r) return;
int mid = (l + r)>>1;
buildTree(rt<<1, l, mid);
buildTree(rt<<1|1, mid + 1, r);
tree[rt] = tree[rt<<1] + tree[rt<<1|1];
}
int query(int rt, int l, int r, int kth)
{
if(l == r) return l;
int mid = (l + r)>>1;
if(tree[rt<<1] >= kth) return query(rt<<1, l, mid, kth);
else return query(rt<<1|1, mid + 1, r, kth - tree[rt<<1]);
}
void update(int rt, int l, int r, int qx)
{
tree[rt]--;
if(l == r) return;
int mid = (l + r)>>1;
if(qx <= mid) update(rt<<1, l, mid, qx);
else update(rt<<1|1, mid + 1, r, qx);
}
int pre_sum[maxN<<1];
int main()
{
int T; scanf("%d", &T);
pre_did();
while(T--)
{
scanf("%d%d", &N, &K);
if(N > 9)
{
ans[1] = N; ans[2] = 1;
memset(vis, false, sizeof(vis));
vis[1] = vis
 = true;
printf("%d %d", N, 1);
buildTree(1, 1, N);
update(1, 1, N, 1); update(1, 1, N, N);
for(int i=1; i<=N-1; i++) pre_sum[i] = pre_sum[i-1] + i;
for(int i=1; i<=N-1; i++) pre_sum[i + N - 1] = pre_sum[i + N - 2] + N - i;
ll Big_pos = K / jiecheng[N - 2] + (K % jiecheng[N - 2] == 0 ? 0 : 1); //大区间
int pos = (int)(lower_bound(pre_sum + 1, pre_sum + 2 * N - 2, Big_pos) - pre_sum);
K -= pre_sum[pos - 1] * jiecheng[N - 2];
int rang_len;
if(pos < N) rang_len = N - pos;
else rang_len = pos - N + 1;
ll small_pos = K / jiecheng[N - 2];
if(small_pos * jiecheng[N-2] == K) small_pos --;
small_pos++;
K -= (small_pos - 1) * jiecheng[N-2];
for(int i=N-3; i>=0; i--)
{
int tmp = K / jiecheng[i] - (K % jiecheng[i] == 0 ? 1 : 0), now;
//tmp = min(tmp, tree[1] - 1);
printf(" %d", (now = query(1, 1, N, tmp + 1)));
update(1, 1, N, now);
if(K > tmp * jiecheng[i]) K -= tmp * jiecheng[i];
}
printf("\n");
//for(int i=1; i<=N; i++) printf("%d%c", ans[i], i == N ? '\n' : ' ');
}
else
{
for(int i=1; i<=N; i++) printf("%d%c", pre
[K][i], i == N ? '\n' : ' ');
}
}
return 0;
}