CSUST 8.3 早训

其他 2019-08-03 20:54:40 阅读次数: 0

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=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
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(0);cout.tie(0)
# 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 + 7;
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 = 0;
    for(int i = 1;i <= n ;i ++) cin >>a[i],sum += a[i];
        sort(a + 1 ,a + n  + 1);
    int ans = 0;
    while ( n * m > sum){
        ans ++;
        int j = 0;
        
                //b[i] = a[i];
                // if(a[i] && j == 0){
                //     j ++;
                //     a[i] --;
                //     sum ++;
                // }
                // else if(a[i] == n){
                //     sum ++;
                // }
        for(int i = 1;i <= n ; i++) b[i] = a[i];
            if(b[1] < m) {
                sum ++;
                b[1]++;
            }
            for(int i = 2;i <= n;i ++){
                if(a[i] == a[i-1] || a[i] == m) continue;
                b[i] ++;
                sum ++;
            }
            sort(b + 1, b + n + 1);
            for(int i = 1;i <= n ;i ++) a[i] = b[i];


        }


    cout << ans << endl;
    return 0;
}
View Code

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=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
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(0);cout.tie(0)
# 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 + 7;
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[11];

int main(int argc, char const *argv[])
{
    int n;
    cin >> n;
    for(int i = 1;i <= n ; i ++){
        cin >> nod[i].str >> nod[i].a >> nod[i].b;
    }
    for(int i = 1;i <= n;i ++){
     
            nod[i].f1 = nod[i].str % 10;
            nod[i].f2 = (nod[i].str / 10 ) % 10;
            nod[i].f3 =  (nod[i].str / 100 ) % 10;
            nod[i].f4 =  (nod[i].str / 1000 ) % 10;
        
    }
    int flag = 0;
    int sum = 0;
    string str;
    for(int i = 0;i <= 9;i ++){

        for(int j = 0;j <= 9;j ++){

            for(int k = 0;k <= 9;k ++){
                for(int p = 0;p <= 9;p ++){
                    int ans = 0;
                    if(i != j && i != k && i != p && j != k && j != p && k != p)
                    for(int q = 1;q <= n; q ++){
                        int a = 0, b = 0;
                        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 + '0');
                        str.push_back(k + '0');
                        str.push_back(j + '0');
                        str.push_back(i + '0');

                        //sum = p * 1000 + k * 100 + j * 10 + i;
                        flag ++;
                       // break;
                    }
                }
            }
        }
    }
    if(flag == 1) cout << str << endl;
    else if(flag > 1)cout << "Need more data" << endl;
    else cout << "Incorrect data" << endl;
    return 0;
}
View Code

题意


看这个数的数据类型（都是整树，且为正数）。

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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=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
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(0);cout.tie(0)
# 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 + 7;
int n  , m ;

int a[maxn];
//map<int,int> mp;
int b[maxn];

int main()
{
    long double n;
    cin>>n;
    if(n<=127)
    {
        cout<<"byte"<<endl;
    }
    else if(n<=32767)
    {
        cout<<"short"<<endl;
    }
    else if(n<=2147483647)
    {
        cout<<"int"<<endl;
    }
    else if(n<=9223372036854775807)
    {
        cout<<"long"<<endl;
    }
    else
    {
        cout<<"BigInteger"<<endl;
    }
    return 0;
}
View Code

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=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
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(0);cout.tie(0)
# 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 + 7;
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 =1;i <=  n;i ++ ){
        cin >> a[i];
    }
    int ans = 0;
    for(int i = 1;i <= n ;i ++){
        int sum = 0;
        int l = i - 1, r = i + 1;
        if(i == 1) 
        while(a[r] <= a[r - 1] && r != n + 1) sum ++,r ++;
        else if(i == n)
            while(a[l] <= a[l + 1] && l != 0) sum ++, l --;
        else {
            while(a[r] <= a[r - 1] && r != n + 1) sum ++,r ++;
            while(a[l] <= a[l + 1] && l != 0) sum ++,l --;
        }
        ans = max(ans,sum + 1);
    }
    cout << ans << endl;
    return 0;
}
View Code

题意

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=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
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(0);cout.tie(0)
# 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 + 7;
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 = 0;
        for(int i = 0; i < str.size() ; i ++){
            if(str[i] == '\\'){
                ans ++;
                if(ans == 2){
                    rt = str.substr(0,i);
                }
                else if(ans > 2) folder[rt].insert(str.substr(0,i));
            }
          
        }
         files[rt].insert(str);
    }
    int maxx1 = 0,maxx2 = 0;
        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 0;
}
View Code

 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=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
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(0);cout.tie(0)
# 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 + 7;
int n  , m ;

int a[maxn];
//map<int,int> mp;
int b[maxn];
int cnt,prime[N],p[N];
void isprime()
{
    cnt = 0;
    memset(prime,true,sizeof(prime));
    for(int i=2; i<N; i++)
    {
        if(prime[i])
        {
            p[cnt++] = i;
            for(int j=i+i; j<N; j+=i)
                prime[j] = false;
        }
    }
}
long long sum[100];
int main(int argc, char const *argv[])
{
    int n ;
    isprime();
    cin >> n ;
    int flag = 1;
    if(n == 2)flag = 0;
    sum[0] = 1;
    for(int i = 1;i <= 50 ; i++){
        sum[i] = p[i-1];
    }
    if(flag){
        cout << 15 << endl << 10 << endl << 6 << endl;
        for(int i = 4 ;i <= n; i ++) cout << i * 6 << endl;
    }

        else 
        cout << -1 << endl;

        return 0;
    }
View Code

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 + 7);
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 = 0;
    Arrays.fill(prime,1);
    for(int i=2; i< maxn; i++)
    {
        if(prime[i] == 1)
        {
            p[cnt++] = i;
            for(int j=i+i; j<maxn; j+=i)
                prime[j] = 0;
        }
    }
}
 
private Object sizeof(int[] prime) {
    // TODO Auto-generated method stub
    return null;
}
static BigInteger sum[] = new BigInteger[100];
public static void  main(String[] args) {
 
     // int n ;
      isprime();
      Scanner a = new Scanner(System.in);
      int n = a.nextInt();
        int flag = 1;
        if(n == 2)flag = 0;
        //sum[0] = 1;
        for(int i = 1;i <= 50 ; i++){
          //  Object[] p;
            sum[i] = BigInteger.valueOf(p[i-1]);
        }
        if(flag == 1)
        for(int i = 1;i <= n ;i ++){
            BigInteger ans = BigInteger.valueOf(1);
            for(int j = 1;j <= n;j ++){
                if(i!=j) ans = ans.multiply(sum[j]);
            }
            if (flag == 1) {
                System.out.println(ans);
            }
            
        }else System.out.print(-1);
 
}
}
View Code

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