A - Settlers' Training

CodeForces - 63B

题意

给你一串数字，相同的数字为一组，每次可以给一组中的一个数字加一，问这一串数字全变成K需要多少步？

题解

模拟

C++代码

/**

/*@author Victor

/*language C++

*/

//#include <bits/stdc++.h>

#include<iostream>

#include<algorithm>

#include<cstdlib>

#include<cstring>

#include<cstdio>

#include<string>

#include<vector>

#include<bitset>

#include<queue>

#include<deque>

#include<stack>

#include<cmath>

#include<list>

//#include<map>

#include<set>

//#define DEBUG

#define RI register int

using namespace std;

typedef long long ll;

//typedef __int128 lll;

const int N=+;

const int MOD=1e9+;

const double PI = acos(-1.0);

const double EXP = 1E-;

const int INF = 0x3f3f3f3f;

#define pii pair<int,int>

#define pll pair<ll,ll>

#define pil pair<int , ll>

#define pli pair<ll,int>

#define pdl pair<double,ll>

#define pld pair<ll,double>

#define pdd pair<double,double>

#define iput(n) scanf("%d",&n)

#define iiput(a,n) scanf("%d%d",&a,&n)

#define iiiput(a,b,c) scanf("%d%d%d",&a,&b,&c)

#define dput(n) scanf("%lf",&n)

#define llput(n) scanf("%lld",&n)

#define cput(n) scanf("%s",n)

#define puti(n) printf("%d\n",n)

#define putll(n) printf("%lld\n",n)

#define putd(n) printf("%lfd\n",n)

#define _cls(n) memset(n,0,sizeof(n))

#define __cls(n) memset(n,0x3f,sizeof(n))

#define lc rt << 1

#define rc rt <<1|1

#define debug(x) cout << "[ " << x << " ]" << endl

//priority_queue <int,vector<int>,greater<int> > Q;//优先队列递增

//priority_queue<int>Q;//递减

//map<ll,ll>mp;

//set<ll>st;

//stack<>st;

//queue<>Q;

#define F first

#define S second

#define pb push_back

#define PB push_back

#define MP make_pair

#define ALL(x) begin(x), end(x)

#define SZ(x) (int)(x).size()

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

//加速输入挂

# define IOS ios::sync_with_stdio(false); cin.tie();cout.tie()

# define FOR(i,a,n) for(int i=a; i<=n; ++i)

//求二进制中1的个数

//__builtin_popcount(n);

//求2^k

//#define (ll)Pow(2,k) (1LL<<k)

#define to_1(n) __builtin_popcount(n)

//树状数组

#define lowbit(x) (x&-x)

//#ifdef DEBUG

#define fin freopen("input.in", "r", stdin)

#define fout freopen("output.out", "w", stdout);

//#endif

//手动扩栈

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

const int maxn = 2e5 + ;

int n  , m ;

int a[maxn];

//map<int,int> mp;

int b[maxn];

int main(int argc, char const *argv[])

{

    cin >> n >> m;  int sum = ;

    for(int i = ;i <= n ;i ++) cin >>a[i],sum += a[i];

        sort(a +  ,a + n  + );

    int ans = ;

    while ( n * m > sum){

        ans ++;

        int j = ;

                //b[i] = a[i];

                // if(a[i] && j == 0){

                //     j ++;

                //     a[i] --;

                //     sum ++;

                // }

                // else if(a[i] == n){

                //     sum ++;

                // }

        for(int i = ;i <= n ; i++) b[i] = a[i];

            if(b[] < m) {

                sum ++;

                b[]++;

            }

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

                if(a[i] == a[i-] || a[i] == m) continue;

                b[i] ++;

                sum ++;

            }

            sort(b + , b + n + );

            for(int i = ;i <= n ;i ++) a[i] = b[i];

        }

    cout << ans << endl;

    return ;

}

B - Bulls and Cows

CodeForces - 63C

【题意】

给你一个长度为4的数字序列(每个数字都在0~9之间,且不重复出现)
现在让你猜这个长度为4的序列是什么.
猜了之后对方会告诉有几个数字是位置和数字都正确的(猜的数字序列有顺序)
以及有几个数字是数字出现了但是位置不正确.
即给你两个反馈。
现在给你n个猜的过程(猜的4个数字以及对应的两个反馈)
问你是否能唯一的确定一个4位数字的答案

【题解】

列举出来最后的4位数的所有可能。
对于每一种可能 将其对这n个猜的过程验证一遍》
如果验证通过,那么递增sum
最后如果sum==1则说明存在唯一的一个可能.
sum==0说明没有任何一个可能,则数据有错
sum>1则说明需要更多的数据才能确定是哪一个.

C++代码

/**

/*@author Victor

/*language C++

*/

//#include <bits/stdc++.h>

#include<iostream>

#include<algorithm>

#include<cstdlib>

#include<cstring>

#include<cstdio>

#include<string>

#include<vector>

#include<bitset>

#include<queue>

#include<deque>

#include<stack>

#include<cmath>

#include<list>

//#include<map>

#include<set>

//#define DEBUG

#define RI register int

using namespace std;

typedef long long ll;

//typedef __int128 lll;

const int N=+;

const int MOD=1e9+;

const double PI = acos(-1.0);

const double EXP = 1E-;

const int INF = 0x3f3f3f3f;

#define pii pair<int,int>

#define pll pair<ll,ll>

#define pil pair<int , ll>

#define pli pair<ll,int>

#define pdl pair<double,ll>

#define pld pair<ll,double>

#define pdd pair<double,double>

#define iput(n) scanf("%d",&n)

#define iiput(a,n) scanf("%d%d",&a,&n)

#define iiiput(a,b,c) scanf("%d%d%d",&a,&b,&c)

#define dput(n) scanf("%lf",&n)

#define llput(n) scanf("%lld",&n)

#define cput(n) scanf("%s",n)

#define puti(n) printf("%d\n",n)

#define putll(n) printf("%lld\n",n)

#define putd(n) printf("%lfd\n",n)

#define _cls(n) memset(n,0,sizeof(n))

#define __cls(n) memset(n,0x3f,sizeof(n))

#define lc rt << 1

#define rc rt <<1|1

#define debug(x) cout << "[ " << x << " ]" << endl

//priority_queue <int,vector<int>,greater<int> > Q;//优先队列递增

//priority_queue<int>Q;//递减

//map<ll,ll>mp;

//set<ll>st;

//stack<>st;

//queue<>Q;

#define F first

#define S second

#define pb push_back

#define PB push_back

#define MP make_pair

#define ALL(x) begin(x), end(x)

#define SZ(x) (int)(x).size()

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

//加速输入挂

# define IOS ios::sync_with_stdio(false); cin.tie();cout.tie()

# define FOR(i,a,n) for(int i=a; i<=n; ++i)

//求二进制中1的个数

//__builtin_popcount(n);

//求2^k

//#define (ll)Pow(2,k) (1LL<<k)

#define to_1(n) __builtin_popcount(n)

//树状数组

#define lowbit(x) (x&-x)

//#ifdef DEBUG

#define fin freopen("input.in", "r", stdin)

#define fout freopen("output.out", "w", stdout);

//#endif

//手动扩栈

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

const int maxn = 2e5 + ;

int n  , m ;

int a[maxn];

//map<int,int> mp;

int b[maxn];

struct node

{

    int str;

    int a ,  b;

    int f1,f2,f3,f4;

}nod[];

int main(int argc, char const *argv[])

{

    int n;

    cin >> n;

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

        cin >> nod[i].str >> nod[i].a >> nod[i].b;

    }

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

            nod[i].f1 = nod[i].str % ;

            nod[i].f2 = (nod[i].str /  ) % ;

            nod[i].f3 =  (nod[i].str /  ) % ;

            nod[i].f4 =  (nod[i].str /  ) % ;

    }

    int flag = ;

    int sum = ;

    string str;

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

        for(int j = ;j <= ;j ++){

            for(int k = ;k <= ;k ++){

                for(int p = ;p <= ;p ++){

                    int ans = ;

                    if(i != j && i != k && i != p && j != k && j != p && k != p)

                    for(int q = ;q <= n; q ++){

                        int a = , b = ;

                        if(nod[q].f1 == i) {

                            a ++;

                        }

                        if(nod[q].f2 == j) a ++;

                        if(nod[q].f3 == k)a ++;

                        if(nod[q].f4 == p) a ++;

                        if(nod[q].f1 == j || nod[q].f1 ==k || nod[q].f1 == p) b++;

                        if(nod[q].f2 == i || nod[q].f2 ==k || nod[q].f2 == p) b++;

                        if(nod[q].f3 == j || nod[q].f3 ==i || nod[q].f3 == p) b++;

                        if(nod[q].f4 == j || nod[q].f4 ==k || nod[q].f4 == i) b++;

                        if(a == nod[q].a && b == nod[q].b) ans ++;

                    }

                    if(ans == n) {

                        str.push_back(p + '');

                        str.push_back(k + '');

                        str.push_back(j + '');

                        str.push_back(i + '');

                        //sum = p * 1000 + k * 100 + j * 10 + i;

                        flag ++;

                       // break;

                    }

                }

            }

        }

    }

    if(flag == ) cout << str << endl;

    else if(flag > )cout << "Need more data" << endl;

    else cout << "Incorrect data" << endl;

    return ;

}

题意

看这个数的数据类型（都是整树，且为正数）。

解题思路

首先判断长度。 
当长度和极值的长度相同时，在用strcmp判断。

C++代码

/**

/*@author Victor

/*language C++

*/

//#include <bits/stdc++.h>

#include<iostream>

#include<algorithm>

#include<cstdlib>

#include<cstring>

#include<cstdio>

#include<string>

#include<vector>

#include<bitset>

#include<queue>

#include<deque>

#include<stack>

#include<cmath>

#include<list>

//#include<map>

#include<set>

//#define DEBUG

#define RI register int

using namespace std;

typedef long long ll;

//typedef __int128 lll;

const int N=+;

const int MOD=1e9+;

const double PI = acos(-1.0);

const double EXP = 1E-;

const int INF = 0x3f3f3f3f;

#define pii pair<int,int>

#define pll pair<ll,ll>

#define pil pair<int , ll>

#define pli pair<ll,int>

#define pdl pair<double,ll>

#define pld pair<ll,double>

#define pdd pair<double,double>

#define iput(n) scanf("%d",&n)

#define iiput(a,n) scanf("%d%d",&a,&n)

#define iiiput(a,b,c) scanf("%d%d%d",&a,&b,&c)

#define dput(n) scanf("%lf",&n)

#define llput(n) scanf("%lld",&n)

#define cput(n) scanf("%s",n)

#define puti(n) printf("%d\n",n)

#define putll(n) printf("%lld\n",n)

#define putd(n) printf("%lfd\n",n)

#define _cls(n) memset(n,0,sizeof(n))

#define __cls(n) memset(n,0x3f,sizeof(n))

#define lc rt << 1

#define rc rt <<1|1

#define debug(x) cout << "[ " << x << " ]" << endl

//priority_queue <int,vector<int>,greater<int> > Q;//优先队列递增

//priority_queue<int>Q;//递减

//map<ll,ll>mp;

//set<ll>st;

//stack<>st;

//queue<>Q;

#define F first

#define S second

#define pb push_back

#define PB push_back

#define MP make_pair

#define ALL(x) begin(x), end(x)

#define SZ(x) (int)(x).size()

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

//加速输入挂

# define IOS ios::sync_with_stdio(false); cin.tie();cout.tie()

# define FOR(i,a,n) for(int i=a; i<=n; ++i)

//求二进制中1的个数

//__builtin_popcount(n);

//求2^k

//#define (ll)Pow(2,k) (1LL<<k)

#define to_1(n) __builtin_popcount(n)

//树状数组

#define lowbit(x) (x&-x)

//#ifdef DEBUG

#define fin freopen("input.in", "r", stdin)

#define fout freopen("output.out", "w", stdout);

//#endif

//手动扩栈

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

const int maxn = 2e5 + ;

int n  , m ;

int a[maxn];

//map<int,int> mp;

int b[maxn];

int main()

{

    long double n;

    cin>>n;

    if(n<=)

    {

        cout<<"byte"<<endl;

    }

    else if(n<=)

    {

        cout<<"short"<<endl;

    }

    else if(n<=)

    {

        cout<<"int"<<endl;

    }

    else if(n<=)

    {

        cout<<"long"<<endl;

    }

    else

    {

        cout<<"BigInteger"<<endl;

    }

    return ;

}

题意

雨可以向两边蔓延，如果往边上走的高度是非递增的。

解题思路

暴力，O（n^2）。

C++代码

/**

/*@author Victor

/*language C++

*/

//#include <bits/stdc++.h>

#include<iostream>

#include<algorithm>

#include<cstdlib>

#include<cstring>

#include<cstdio>

#include<string>

#include<vector>

#include<bitset>

#include<queue>

#include<deque>

#include<stack>

#include<cmath>

#include<list>

//#include<map>

#include<set>

//#define DEBUG

#define RI register int

using namespace std;

typedef long long ll;

//typedef __int128 lll;

const int N=+;

const int MOD=1e9+;

const double PI = acos(-1.0);

const double EXP = 1E-;

const int INF = 0x3f3f3f3f;

#define pii pair<int,int>

#define pll pair<ll,ll>

#define pil pair<int , ll>

#define pli pair<ll,int>

#define pdl pair<double,ll>

#define pld pair<ll,double>

#define pdd pair<double,double>

#define iput(n) scanf("%d",&n)

#define iiput(a,n) scanf("%d%d",&a,&n)

#define iiiput(a,b,c) scanf("%d%d%d",&a,&b,&c)

#define dput(n) scanf("%lf",&n)

#define llput(n) scanf("%lld",&n)

#define cput(n) scanf("%s",n)

#define puti(n) printf("%d\n",n)

#define putll(n) printf("%lld\n",n)

#define putd(n) printf("%lfd\n",n)

#define _cls(n) memset(n,0,sizeof(n))

#define __cls(n) memset(n,0x3f,sizeof(n))

#define lc rt << 1

#define rc rt <<1|1

#define debug(x) cout << "[ " << x << " ]" << endl

//priority_queue <int,vector<int>,greater<int> > Q;//??????

//priority_queue<int>Q;//??

//map<ll,ll>mp;

//set<ll>st;

//stack<>st;

//queue<>Q;

#define F first

#define S second

#define pb push_back

#define PB push_back

#define MP make_pair

#define ALL(x) begin(x), end(x)

#define SZ(x) (int)(x).size()

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

//?????

# define IOS ios::sync_with_stdio(false); cin.tie();cout.tie()

# define FOR(i,a,n) for(int i=a; i<=n; ++i)

//?????1???

//__builtin_popcount(n);

//?2^k

//#define (ll)Pow(2,k) (1LL<<k)

#define to_1(n) __builtin_popcount(n)

//????

#define lowbit(x) (x&-x)

//#ifdef DEBUG

#define fin freopen("input.in", "r", stdin)

#define fout freopen("output.out", "w", stdout);

//#endif

//????

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

const int maxn = 2e5 + ;

int n  , m ;

int a[maxn];

//map<int,int> mp;

int b[maxn];

int main(int argc, char const *argv[])

{

    int n ;

    cin >> n ;

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

        cin >> a[i];

    }

    int ans = ;

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

        int sum = ;

        int l = i - , r = i + ;

        if(i == )

        while(a[r] <= a[r - ] && r != n + ) sum ++,r ++;

        else if(i == n)

            while(a[l] <= a[l + ] && l != ) sum ++, l --;

        else {

            while(a[r] <= a[r - ] && r != n + ) sum ++,r ++;

            while(a[l] <= a[l + ] && l != ) sum ++,l --;

        }

        ans = max(ans,sum + );

    }

    cout << ans << endl;

    return ;

}
给若干个文件路径，问子文件夹以及子文件最多为多少 模拟题，可以直接用文件的绝对路径来表示一个文件 

题解

这样就不用担心不同路径文件的重名问题，也不用建树了 
其次子文件夹和子文件数最多的肯定是根目录下的文件夹 
用set维护一下这些文件夹的情况即可，set的好处是不用去重

C++代码

/**

/*@author Victor

/*language C++

*/

//#include <bits/stdc++.h>

#include<iostream>

#include<algorithm>

#include<cstdlib>

#include<cstring>

#include<cstdio>

#include<string>

#include<vector>

#include<bitset>

#include<queue>

#include<deque>

#include<stack>

#include<cmath>

#include<list>

#include<map>

#include<set>

//#define DEBUG

#define RI register int

using namespace std;

typedef long long ll;

//typedef __int128 lll;

const int N=+;

const int MOD=1e9+;

const double PI = acos(-1.0);

const double EXP = 1E-;

const int INF = 0x3f3f3f3f;

#define pii pair<int,int>

#define pll pair<ll,ll>

#define pil pair<int , ll>

#define pli pair<ll,int>

#define pdl pair<double,ll>

#define pld pair<ll,double>

#define pdd pair<double,double>

#define iput(n) scanf("%d",&n)

#define iiput(a,n) scanf("%d%d",&a,&n)

#define iiiput(a,b,c) scanf("%d%d%d",&a,&b,&c)

#define dput(n) scanf("%lf",&n)

#define llput(n) scanf("%lld",&n)

#define cput(n) scanf("%s",n)

#define puti(n) printf("%d\n",n)

#define putll(n) printf("%lld\n",n)

#define putd(n) printf("%lfd\n",n)

#define _cls(n) memset(n,0,sizeof(n))

#define __cls(n) memset(n,0x3f,sizeof(n))

#define lc rt << 1

#define rc rt <<1|1

#define debug(x) cout << "[ " << x << " ]" << endl

//priority_queue <int,vector<int>,greater<int> > Q;//优先队列递增

//priority_queue<int>Q;//递减

//map<ll,ll>mp;

//set<ll>st;

//stack<>st;

//queue<>Q;

#define F first

#define S second

#define pb push_back

#define PB push_back

#define MP make_pair

#define ALL(x) begin(x), end(x)

#define SZ(x) (int)(x).size()

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

//加速输入挂

# define IOS ios::sync_with_stdio(false); cin.tie();cout.tie()

# define FOR(i,a,n) for(int i=a; i<=n; ++i)

//求二进制中1的个数

//__builtin_popcount(n);

//求2^k

//#define (ll)Pow(2,k) (1LL<<k)

#define to_1(n) __builtin_popcount(n)

//树状数组

#define lowbit(x) (x&-x)

//#ifdef DEBUG

#define fin freopen("input.in", "r", stdin)

#define fout freopen("output.out", "w", stdout);

//#endif

//手动扩栈

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

const int maxn = 2e5 + ;

int n  , m ;

int a[maxn];

//map<int,int> mp;

int b[maxn];

map<string,set<string> > folder,files;

int main(int argc, char const *argv[])

{

    string str , rt;

    while(cin >> str){

        int ans = ;

        for(int i = ; i < str.size() ; i ++){

            if(str[i] == '\\'){

                ans ++;

                if(ans == ){

                    rt = str.substr(,i);

                }

                else if(ans > ) folder[rt].insert(str.substr(,i));

            }

        }

         files[rt].insert(str);

    }

    int maxx1 = ,maxx2 = ;

        for(auto au : folder){

            maxx1 = max(maxx1,(int)au.second.size());

        }

        for(auto au : files){

            maxx2 = max(maxx2,(int)au.second.size());

        }

        cout << maxx1 << " " << maxx2 << endl;

    return ;

}

题解

要求构造一个长度为n的正整数序列ai，使得对于∀i≠j,gcd(ai,aj)>1，且gcd(a1,a2,...,an)=1
n=2显然无解，n>2时，取前nn个素数p1,p2,...,pn令ai=∏j≠ipj，简单验证知a1,a2,...,an满足条件，数字很大要用到高精度

解法二

构造出前3个15 10 6 剩下的为 i * 6即可

C++代码一

/**

/*@author Victor

/*language C++

*/

//#include <bits/stdc++.h>

#include<iostream>

#include<algorithm>

#include<cstdlib>

#include<cstring>

#include<cstdio>

#include<string>

#include<vector>

#include<bitset>

#include<queue>

#include<deque>

#include<stack>

#include<cmath>

#include<list>

#include<map>

#include<set>

//#define DEBUG

#define RI register int

using namespace std;

typedef long long ll;

//typedef __int128 lll;

const int N=+;

const int MOD=1e9+;

const double PI = acos(-1.0);

const double EXP = 1E-;

const int INF = 0x3f3f3f3f;

#define pii pair<int,int>

#define pll pair<ll,ll>

#define pil pair<int , ll>

#define pli pair<ll,int>

#define pdl pair<double,ll>

#define pld pair<ll,double>

#define pdd pair<double,double>

#define iput(n) scanf("%d",&n)

#define iiput(a,n) scanf("%d%d",&a,&n)

#define iiiput(a,b,c) scanf("%d%d%d",&a,&b,&c)

#define dput(n) scanf("%lf",&n)

#define llput(n) scanf("%lld",&n)

#define cput(n) scanf("%s",n)

#define puti(n) printf("%d\n",n)

#define putll(n) printf("%lld\n",n)

#define putd(n) printf("%lfd\n",n)

#define _cls(n) memset(n,0,sizeof(n))

#define __cls(n) memset(n,0x3f,sizeof(n))

#define lc rt << 1

#define rc rt <<1|1

#define debug(x) cout << "[ " << x << " ]" << endl

//priority_queue <int,vector<int>,greater<int> > Q;//优先队列递增

//priority_queue<int>Q;//递减

//map<ll,ll>mp;

//set<ll>st;

//stack<>st;

//queue<>Q;

#define F first

#define S second

#define pb push_back

#define PB push_back

#define MP make_pair

#define ALL(x) begin(x), end(x)

#define SZ(x) (int)(x).size()

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

//加速输入挂

# define IOS ios::sync_with_stdio(false); cin.tie();cout.tie()

# define FOR(i,a,n) for(int i=a; i<=n; ++i)

//求二进制中1的个数

//__builtin_popcount(n);

//求2^k

//#define (ll)Pow(2,k) (1LL<<k)

#define to_1(n) __builtin_popcount(n)

//树状数组

#define lowbit(x) (x&-x)

//#ifdef DEBUG

#define fin freopen("input.in", "r", stdin)

#define fout freopen("output.out", "w", stdout);

//#endif

//手动扩栈

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

const int maxn = 2e5 + ;

int n  , m ;

int a[maxn];

//map<int,int> mp;

int b[maxn];

int cnt,prime[N],p[N];

void isprime()

{

    cnt = ;

    memset(prime,true,sizeof(prime));

    for(int i=; i<N; i++)

    {

        if(prime[i])

        {

            p[cnt++] = i;

            for(int j=i+i; j<N; j+=i)

                prime[j] = false;

        }

    }

}

long long sum[];

int main(int argc, char const *argv[])

{

    int n ;

    isprime();

    cin >> n ;

    int flag = ;

    if(n == )flag = ;

    sum[] = ;

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

        sum[i] = p[i-];

    }

    if(flag){

        cout <<  << endl <<  << endl <<  << endl;

        for(int i =  ;i <= n; i ++) cout << i *  << endl;

    }

        else

        cout << - << endl;

        return ;

    }

java代码二

 //package ce;

import java.lang.invoke.ConstantCallSite;

import java.math.BigInteger;

import java.util.Arrays;

import java.util.Scanner;

import javax.swing.InputMap;

import javax.swing.undo.AbstractUndoableEdit;

public class Main{

static int maxn = (int) (1e5 + );

static int p[] = new int [maxn];

static int prime[] = new int [maxn];

static void isprime()

{

    //int b[] = new int[maxn];

    int cnt;

   // p[] = new int [maxn];

    cnt = ;

    Arrays.fill(prime,);

    for(int i=; i< maxn; i++)

    {

        if(prime[i] == )

        {

            p[cnt++] = i;

            for(int j=i+i; j<maxn; j+=i)

                prime[j] = ;

        }

    }

}

private Object sizeof(int[] prime) {

    // TODO Auto-generated method stub

    return null;

}

static BigInteger sum[] = new BigInteger[];

public static void  main(String[] args) {

     // int n ;

      isprime();

      Scanner a = new Scanner(System.in);

      int n = a.nextInt();

        int flag = ;

        if(n == )flag = ;

        //sum[0] = 1;

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

          //  Object[] p;

            sum[i] = BigInteger.valueOf(p[i-]);

        }

        if(flag == )

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

            BigInteger ans = BigInteger.valueOf();

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

                if(i!=j) ans = ans.multiply(sum[j]);

            }

            if (flag == ) {

                System.out.println(ans);

            }

        }else System.out.print(-);

}

}

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