http://acm.hdu.edu.cn/showproblem.php?pid=4850

題意：構造長度為n的字元序列，使得>=4的子串只出現一次

其實最長只能構造出來26^4+4-1= 456979 的序列，大於該數的都是不可能的。構造方法，就是那種歐拉回路的序列，此題DFS會爆棧，手動擴充套件棧也可以AC......

遞迴形式的開始WA了，沒有細調就換非遞迴了，後來又想了想，雖然自己電腦上執行不了，但是先把長度按小的來，然後除錯程式碼，然後在擴大，AC了，當時錯在MOD，遞迴的MOD應該是26^4，而不是26^4+1，因為控制在0~（26^4-1）範圍內，就是456976個數

所以要變成非遞迴，我其實不太理解非遞迴的，然後參考自己以前做過的也是不太理解的這個程式碼http://blog.csdn.net/u011026968/article/details/38151303

然後AC

非遞迴版  46ms AC

//#pragma comment(linker, "/STACK:102400000,102400000")
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <iomanip>
#include <cmath>
#include <map>
#include <set>
#include <queue>
using namespace std;

#define ls(rt) rt*2
#define rs(rt) rt*2+1
#define ll long long
#define ull unsigned long long
#define rep(i,s,e) for(int i=s;i<e;i++)
#define repe(i,s,e) for(int i=s;i<=e;i++)
#define CL(a,b) memset(a,b,sizeof(a))
#define IN(s) freopen(s,"r",stdin)
#define OUT(s) freopen(s,"w",stdout)
const ll ll_INF = ((ull)(-1))>>1;
const double EPS = 1e-8;
const int INF = 100000000;
const int MAXN = 101;
const int S = 500000;
const int SIZE = 26;
const int LEN = 456976+3;
//const int MOD = 456976;
const int MOD =17576;

char str[LEN+10];
char li[LEN*SIZE+10];
int sta[LEN*SIZE+10];

/*int dfs(int cnt, int s)
{
    //printf("cnt=%d %d\n",cnt,s);
    if(cnt == LEN)return 1;
    for(int i=0;i<SIZE;i++)
    {
        if(!vis[(s*SIZE+i)%MOD])///
        {
            vis[(s*SIZE+i)%MOD]=1;
            str[cnt]=i;
            if(dfs(cnt+1, (s*SIZE+i)%MOD))return 1;
            //vis[(s<<1)+i]=0;
        }
    }
    return 0;
}*/
int tp,ans;
void sea(int v)
{
    while(li[v]<SIZE)
    {
        int w=v*SIZE+li[v];
        li[v]++;
        sta[tp++]=w;
        v=w%MOD;
    }
}
void solve()
{
    CL(str,0);
    CL(li,0);
    tp=0;
    int v;
    sea(0);
    str[0]='a';
    ans=1;
    while(tp)
    {
        v=sta[--tp];
        str[ans++]=v%SIZE+'a';
        v/=SIZE;
        sea(v);
    }
}

int main()
{
    //OUT("hdu4850.txt");
    solve();
    int n;
    while(~scanf("%d",&n))
    {
        if(n>LEN)puts("Impossible");
        else
        {
            if(n<=4)
            {
                for(int i=0;i<n;i++)
                    putchar('a');
            }
            else
            {
                for(int i=1;i<4;i++)putchar('a');
                int tt=ans;
                n-=3;
                while(tt>ans-n)putchar(str[--tt]);
            }
            putchar('\n');
        }
    }
    return 0;
}
 

遞迴版 93ms AC

#pragma comment(linker, "/STACK:102400000,102400000")
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <iomanip>
#include <cmath>
#include <map>
#include <set>
#include <queue>
using namespace std;

#define ls(rt) rt*2
#define rs(rt) rt*2+1
#define ll long long
#define ull unsigned long long
#define rep(i,s,e) for(int i=s;i<e;i++)
#define repe(i,s,e) for(int i=s;i<=e;i++)
#define CL(a,b) memset(a,b,sizeof(a))
#define IN(s) freopen(s,"r",stdin)
#define OUT(s) freopen(s,"w",stdout)
const ll ll_INF = ((ull)(-1))>>1;
const double EPS = 1e-8;
const int INF = 100000000;
const int MAXN = 101;
const int S = 500000;
const int SIZE = 26;
const int LEN = 456976+3;
const int MOD = 456976;

int str[LEN+10];
int vis[LEN+10];

int dfs(int cnt, int s)
{
    //printf("cnt=%d %d\n",cnt,s);
    if(cnt == LEN)return 1;
    for(int i=0;i<SIZE;i++)
    {
        if(!vis[(s*SIZE+i)%MOD])///
        {
            vis[(s*SIZE+i)%MOD]=1;
            str[cnt]=i;
            if(dfs(cnt+1, (s*SIZE+i)%MOD))return 1;
        }
    }
    return 0;
}


void init()
{
    CL(str,0);
    CL(vis,0);
    vis[0]=1;
    dfs(4,0);
}

int main()
{
    //OUT("hdu4850.txt");
    init();
    int n;
    while(~scanf("%d",&n))
    {
        if(n>LEN)puts("Impossible");
        else
        {
            for(int i=0;i<n;i++)
                putchar(str[i]+'a');
            putchar('\n');
        }
    }
    return 0;
}