11 | 6月 | 2018 | 超客的水墨烟雨

uva524

maksyuki 发表于 oj 分类，标签: 基础算法-搜索(DFS\BFS)\记忆化

11 6月 2018

A ring is composed of n (even number) circles as shown in diagram. Put natural numbers 1, 2, . . . , n into each circle separately, and the sum of numbers in two adjacent circles should be a prime. Note: the number of first circle should always be 1.
Input
n (0 < n ≤ 16)
Output
The output format is shown as sample below. Each row represents a series of circle numbers in the ring beginning from 1 clockwisely and anticlockwisely. The order of numbers must satisfy the above requirements.
You are to write a program that completes above process.
Sample Input
6
8
Sample Output
Case 1:
1 4 3 2 5 6
1 6 5 2 3 4
Case 2:
1 2 3 8 5 6 7 4
1 2 5 8 3 4 7 6
1 4 7 6 5 8 3 2
1 6 7 4 3 8 5 2

题目类型：递归枚举
算法分析：从1开始生成排列并在生成过程中判断相邻两个数之和是否是素数

/**************************************************
filename       :f.cpp
author         :maksyuki
created time   :2018/6/11 14:48:25
last modified  :2018/6/11 15:11:57
file location  :C:\Users\abcd\Desktop\TheEternalPoet
***************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("in", "r", stdin)
#define CFO freopen ("out", "w", stdout)
#define CPPFF ifstream cin ("in")
#define CPPFO ofstream cout ("out")
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl
#define	DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC)
#define PB push_back
#define MP(A, B) make_pair(A, B)

typedef long long LL;
typedef unsigned long long ULL;
typedef double DB;
typedef pair <int, int> PII;
typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 66;

int n, vv[maxn];
bool vis[maxn], is_prime[maxn];

bool solve(int v) {
    for(int i = 2; i * i <= v; i++)
        if(v % i == 0) return false;
    return true;
}

void dfs(int p) {
    if(p == n + 1 && is_prime[vv[1]+vv[n]]) {
        for(int i = 1; i <= n; i++) {
            if(i == 1) cout << vv[i];
            else cout << " " << vv[i];
        }
        cout << endl;
        return ;
    }

    for(int i = 2; i <= n; i++) {
        if(!vis[i] && is_prime[i+vv[p-1]]) {
            vis[i] = true;
            vv[p] = i;
            dfs(p + 1);
            vis[i] = false;
        }
    }
}

int main()
{
#ifdef LOCAL
    CFF;
    //CFO;
#endif
    
    int id = 1;
    memset(is_prime, false, sizeof(is_prime));
    for(int i = 1; i < maxn * 2; i++)
        is_prime[i] = solve(i);

    while(cin >> n) {
        if(id > 1) cout << endl;

        cout << "Case " << id++ << ":" << endl;
        memset(vv, 0, sizeof(vv));
        memset(vis, false, sizeof(vis));
        vis[1] = true;
        vv[1] = 1;
        dfs(2);
    }
    return 0;
}

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/**************************************************

filename :f.cpp

author :maksyuki

created time :2018/6/11 14:48:25

last modified :2018/6/11 15:11:57

file location :C:\Users\abcd\Desktop\TheEternalPoet

***************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("in", "r", stdin)

#define CFO freopen ("out", "w", stdout)

#define CPPFF ifstream cin ("in")

#define CPPFO ofstream cout ("out")

#define DB(ccc) cout << #ccc << " = " << ccc << endl

#define DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC)

#define PB push_back

#define MP(A, B) make_pair(A, B)

typedef long long LL;

typedef unsigned long long ULL;

typedef double DB;

typedef pair <int, int> PII;

typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 66;

int n, vv[maxn];

bool vis[maxn], is_prime[maxn];

bool solve(int v) {

for(int i = 2; i * i <= v; i++)

if(v % i == 0) return false;

return true;

}

void dfs(int p) {

if(p == n + 1 && is_prime[vv[1]+vv[n]]) {

for(int i = 1; i <= n; i++) {

if(i == 1) cout << vv[i];

else cout << " " << vv[i];

}

cout << endl;

return ;

}

for(int i = 2; i <= n; i++) {

if(!vis[i] && is_prime[i+vv[p-1]]) {

vis[i] = true;

vv[p] = i;

dfs(p + 1);

vis[i] = false;

}

int main()

{

#ifdef LOCAL

CFF;

//CFO;

#endif

int id = 1;

memset(is_prime, false, sizeof(is_prime));

for(int i = 1; i < maxn * 2; i++)

is_prime[i] = solve(i);

while(cin >> n) {

if(id > 1) cout << endl;

cout << "Case " << id++ << ":" << endl;

memset(vv, 0, sizeof(vv));

memset(vis, false, sizeof(vis));

vis[1] = true;

vv[1] = 1;

dfs(2);

}

return 0;

}

uva10976

maksyuki 发表于 oj 分类，标签: 基础算法-枚举

11 6月 2018

It is easy to see that for every fraction in the form 1/k(k > 0), we can always find two positive integers x and y, x ≥ y, such that: 1/k= 1/x + 1/y
Now our question is: can you write a program that counts how many such pairs of x and y there are for any given k?
Input
Input contains no more than 100 lines, each giving a value of k (0 < k ≤ 10000).
Output
For each k, output the number of corresponding (x, y) pairs, followed by a sorted list of the values of x and y, as shown in the sample output.
Sample Input
2
12
Sample Output
2
1/2 = 1/6 + 1/3
1/2 = 1/4 + 1/4
8
1/12 = 1/156 + 1/13
1/12 = 1/84 + 1/14
1/12 = 1/60 + 1/15
1/12 = 1/48 + 1/16
1/12 = 1/36 + 1/18
1/12 = 1/30 + 1/20
1/12 = 1/28 + 1/21
1/12 = 1/24 + 1/24

题目类型：暴力枚举

算法分析：直接枚举出y，然后再计算出x是否满足要求，由x >= y，则可得1/x <= 1/y，代入表达式可得y <= 2k，所以枚举[1,2k]内的值作为y，然后生成x并判断

/**************************************************
filename       :e.cpp
author         :maksyuki
created time   :2018/6/11 11:06:29
last modified  :2018/6/11 11:14:36
file location  :C:\Users\abcd\Desktop\TheEternalPoet
***************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("in", "r", stdin)
#define CFO freopen ("out", "w", stdout)
#define CPPFF ifstream cin ("in")
#define CPPFO ofstream cout ("out")
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl
#define	DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC)
#define PB push_back
#define MP(A, B) make_pair(A, B)

typedef long long LL;
typedef unsigned long long ULL;
typedef double DB;
typedef pair <int, int> PII;
typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 1e5 + 6666;

typedef pair<long long, long long> PLL;

LL gcd(LL a, LL b) {
    return (!b)? a : gcd(b, a % b);
}

int main()
{
#ifdef LOCAL
    CFF;
    //CFO;
#endif
    
    LL k;
    vector<PLL> ans;
    while(cin >> k) {
        ans.clear();
        for(LL y = 1; y <= 2 * k; y++) {
            LL ta = gcd(y, k);
            LL tb = y * k / ta;
            LL tmpa = tb / k - tb / y;
            LL tmpb = gcd(tmpa, tb);
            tmpa /= tmpb;
            tb /= tmpb;

            if(tmpa == 1 && tb >= y)
                ans.emplace_back(PLL(y, tb));
        }

        int alen = ans.size();
        cout << alen << endl;
        for(int i = 0; i < alen; i++)
            cout << "1/" << k << " = " << "1/" << ans[i].second << " + " << "1/" << ans[i].first << endl;
    }
    return 0;
}

/**************************************************

filename :e.cpp

author :maksyuki

created time :2018/6/11 11:06:29

last modified :2018/6/11 11:14:36

file location :C:\Users\abcd\Desktop\TheEternalPoet

***************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("in", "r", stdin)

#define CFO freopen ("out", "w", stdout)

#define CPPFF ifstream cin ("in")

#define CPPFO ofstream cout ("out")

#define DB(ccc) cout << #ccc << " = " << ccc << endl

#define DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC)

#define PB push_back

#define MP(A, B) make_pair(A, B)

typedef long long LL;

typedef unsigned long long ULL;

typedef double DB;

typedef pair <int, int> PII;

typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 1e5 + 6666;

typedef pair<long long, long long> PLL;

LL gcd(LL a, LL b) {

return (!b)? a : gcd(b, a % b);

}

int main()

{

#ifdef LOCAL

CFF;

//CFO;

#endif

LL k;

vector<PLL> ans;

while(cin >> k) {

ans.clear();

for(LL y = 1; y <= 2 * k; y++) {

LL ta = gcd(y, k);

LL tb = y * k / ta;

LL tmpa = tb / k - tb / y;

LL tmpb = gcd(tmpa, tb);

tmpa /= tmpb;

tb /= tmpb;

if(tmpa == 1 && tb >= y)

ans.emplace_back(PLL(y, tb));

}

int alen = ans.size();

cout << alen << endl;

for(int i = 0; i < alen; i++)

cout << "1/" << k << " = " << "1/" << ans[i].second << " + " << "1/" << ans[i].first << endl;

}

return 0;

}

uva11059

maksyuki 发表于 oj 分类，标签: 基础算法-枚举

11 6月 2018

Given a sequence of integers S = {S1, S2, . . . , Sn}, you should determine what is the value of the maximum positive product involving consecutive terms of S. If you cannot find a positive sequence, you should consider 0 as the value of the maximum product.
Input
Each test case starts with 1 ≤ N ≤ 18, the number of elements in a sequence. Each element Si is an integer such that −10 ≤ Si ≤ 10. Next line will have N integers, representing the value of each element in the sequence. There is a blank line after each test case. The input is terminated by end of file (EOF).
Output
For each test case you must print the message: ‘Case #M: The maximum product is P.’, where M is the number of the test case, starting from 1, and P is the value of the maximum product. After each test case you must print a blank line.
Sample Input
3
2 4 -3
5
2 5 -1 2 -1
Sample Output
Case #1: The maximum product is 8.
Case #2: The maximum product is 20.

题目类型：暴力枚举
算法分析：直接枚举序列的起始和终止位置，然后生成所有的序列并更新最大值，注意需要使用long long类型

/**************************************************
filename       :d.cpp
author         :maksyuki
created time   :2018/6/11 10:37:52
last modified  :2018/6/11 10:55:49
file location  :C:\Users\abcd\Desktop\TheEternalPoet
***************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("in", "r", stdin)
#define CFO freopen ("out", "w", stdout)
#define CPPFF ifstream cin ("in")
#define CPPFO ofstream cout ("out")
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl
#define	DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC)
#define PB push_back
#define MP(A, B) make_pair(A, B)

typedef long long LL;
typedef unsigned long long ULL;
typedef double DB;
typedef pair <int, int> PII;
typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 66;


LL aa[maxn];

int main()
{
#ifdef LOCAL
    CFF;
    //CFO;
#endif
    
    int n;
    int id = 1;
    while(cin >> n) {
        for(int i = 1; i <= n; i++)
            cin >> aa[i];

        LL ans = -INF;
        for(int i = 1; i <= n; i++) {
            LL sum = aa[i];
            ans = max(ans, sum);
            for(int j = i + 1; j <= n; j++) {
                sum *= aa[j];
                ans = max(ans, sum);
            }
        }
        
        cout << "Case #" << id++ << ": The maximum product is ";
        if(ans < 0) cout << "0." << endl;
        else cout << ans << "." << endl;
        cout << endl;
    }
    return 0;
}

/**************************************************

filename :d.cpp

author :maksyuki

created time :2018/6/11 10:37:52

last modified :2018/6/11 10:55:49

file location :C:\Users\abcd\Desktop\TheEternalPoet

***************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("in", "r", stdin)

#define CFO freopen ("out", "w", stdout)

#define CPPFF ifstream cin ("in")

#define CPPFO ofstream cout ("out")

#define DB(ccc) cout << #ccc << " = " << ccc << endl

#define DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC)

#define PB push_back

#define MP(A, B) make_pair(A, B)

typedef long long LL;

typedef unsigned long long ULL;

typedef double DB;

typedef pair <int, int> PII;

typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 66;

LL aa[maxn];

int main()

{

#ifdef LOCAL

CFF;

//CFO;

#endif

int n;

int id = 1;

while(cin >> n) {

for(int i = 1; i <= n; i++)

cin >> aa[i];

LL ans = -INF;

for(int i = 1; i <= n; i++) {

LL sum = aa[i];

ans = max(ans, sum);

for(int j = i + 1; j <= n; j++) {

sum *= aa[j];

ans = max(ans, sum);

}

cout << "Case #" << id++ << ": The maximum product is ";

if(ans < 0) cout << "0." << endl;

else cout << ans << "." << endl;

cout << endl;

}

return 0;

}

uva725

maksyuki 发表于 oj 分类，标签: 基础算法-枚举

11 6月 2018

Write a program that finds and displays all pairs of 5-digit numbers that between them use the digits 0 through 9 once each, such that the first number divided by the second is equal to an integer N, where 2 ≤ N ≤ 79. That is,
abcde / fghij = N, where each letter represents a different digit. The first digit of one of the numerals is allowed to be
zero.
Input
Each line of the input file consists of a valid integer N. An input of zero is to terminate the program.
Output
Your program have to display ALL qualifying pairs of numerals, sorted by increasing numerator (and, of course, denominator). Your output should be in the following general form:
xxxxx / xxxxx = N
xxxxx / xxxxx = N
.
.
In case there are no pairs of numerals satisfying the condition, you must write ‘There are no solutions for N.’. Separate the output for two different values of N by a blank line.
Sample Input
61
62
0
Sample Output
There are no solutions for 61.
79546 / 01283 = 62
94736 / 01528 = 62

题目类型：暴力枚举

算法分析：直接枚举除数并乘以n，看被除数是否满足要求

/**************************************************
filename       :c.cpp
author         :maksyuki
created time   :2018/6/11 0:08:30
last modified  :2018/6/11 9:51:42
file location  :C:\Users\abcd\Desktop\TheEternalPoet
***************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("in", "r", stdin)
#define CFO freopen ("out", "w", stdout)
#define CPPFF ifstream cin ("in")
#define CPPFO ofstream cout ("out")
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl
#define	DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC)
#define PB push_back
#define MP(A, B) make_pair(A, B)

typedef long long LL;
typedef unsigned long long ULL;
typedef double DB;
typedef pair <int, int> PII;
typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 1e5 + 6666;

int n;
bool is_have = false;
bool dig[16];
bool vis[16];

void dfs(int p, string sa) {
    if(p == 6) {
        int v = stoi(sa);
        int tt = v * n;
        string sb = to_string(tt);
        if(sb.size() <= 5) {
            bool is_valid = true;
            for(int i = 0; i < 16; i++)
                vis[i] = dig[i];

            for(int i = 0; sb[i]; i++) {
                if(vis[sb[i]-'0']) {
                    is_valid = false;
                    break;
                }
                else 
                    vis[sb[i]-'0'] = true;
            }

            if(is_valid) {
                is_have = true;
                if(sb.size() == 4) {
                    if(!vis[0])
                        cout << "0" << sb << " / " << sa << " = " << n << endl;
                }

                else
                    cout << sb << " / " << sa << " = " << n << endl;
            }
        }
        return ;
    }
    
    for(int i = 0; i <= 9; i++)
        if(!dig[i]) {
            dig[i] = true;
            dfs(p + 1, sa + to_string(i));
            dig[i] = false;
        }
}

int main()
{
#ifdef LOCAL
    CFF;
    //CFO;
#endif

    bool is_first = true;
    while(cin >> n) {
        if(!n) break;

        is_have = false;
        memset(dig, false, sizeof(dig));

        if(is_first) is_first = false;
        else cout << endl;

        dfs(1, "");
        if(!is_have)
            cout << "There are no solutions for " << n << "." << endl;
    }
    return 0;
}

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/**************************************************

filename :c.cpp

author :maksyuki

created time :2018/6/11 0:08:30

last modified :2018/6/11 9:51:42

file location :C:\Users\abcd\Desktop\TheEternalPoet

***************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("in", "r", stdin)

#define CFO freopen ("out", "w", stdout)

#define CPPFF ifstream cin ("in")

#define CPPFO ofstream cout ("out")

#define DB(ccc) cout << #ccc << " = " << ccc << endl

#define DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC)

#define PB push_back

#define MP(A, B) make_pair(A, B)

typedef long long LL;

typedef unsigned long long ULL;

typedef double DB;

typedef pair <int, int> PII;

typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 1e5 + 6666;

int n;

bool is_have = false;

bool dig[16];

bool vis[16];

void dfs(int p, string sa) {

if(p == 6) {

int v = stoi(sa);

int tt = v * n;

string sb = to_string(tt);

if(sb.size() <= 5) {

bool is_valid = true;

for(int i = 0; i < 16; i++)

vis[i] = dig[i];

for(int i = 0; sb[i]; i++) {

if(vis[sb[i]-'0']) {

is_valid = false;

break;

}

else

vis[sb[i]-'0'] = true;

}

if(is_valid) {

is_have = true;

if(sb.size() == 4) {

if(!vis[0])

cout << "0" << sb << " / " << sa << " = " << n << endl;

}

else

cout << sb << " / " << sa << " = " << n << endl;

}

return ;

}

for(int i = 0; i <= 9; i++)

if(!dig[i]) {

dig[i] = true;

dfs(p + 1, sa + to_string(i));

dig[i] = false;

}

int main()

{

#ifdef LOCAL

CFF;

//CFO;

#endif

bool is_first = true;

while(cin >> n) {

if(!n) break;

is_have = false;

memset(dig, false, sizeof(dig));

if(is_first) is_first = false;

else cout << endl;

dfs(1, "");

if(!is_have)

cout << "There are no solutions for " << n << "." << endl;

}

return 0;

}

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