Educational Codeforces Round 83 (Rated for Div. 2)

A. Two Regular Polygons

You are given two integers n and m (m Examples of convex regular polygons
Your task is to say if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon.
You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1≤t≤104) — the number of test cases.
The next t lines describe test cases. Each test case is given as two space-separated integers n and m (3≤m

Output

For each test case, print the answer — “YES” (without quotes), if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon and “NO” otherwise.

Example

input

2
6 3
7 3

output

YES
NO

题意：两个正多边形的顶点是否能完全重合。

思路：签到。边数为a，b（a>b）满足a%b==0即可。

#include 
using namespace std;
#define pi acos(-1)
#define N 1000000007
#define ll long long
#define ull unsigned long long
#define mem(a) memset(a,0,sizeof(a))

int main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    int t;
    cin >> t;
    while(t--){
        int n, m;
        cin >> n >> m;
        if(n%m==0){
            cout << "YES" << endl;
        }else{
            cout << "NO" << endl;
        }
    }
    return 0;
}

B. Bogosort

You are given an array a1,a2,…,an. Array is good if for each pair of indexes i For example, if a=[1,1,3,5], then shuffled arrays [1,3,5,1], [3,5,1,1] and [5,3,1,1] are good, but shuffled arrays [3,1,5,1], [1,1,3,5] and [1,1,5,3] aren’t.
It’s guaranteed that it’s always possible to shuffle an array to meet this condition.

Input

The first line contains one integer t (1≤t≤100) — the number of test cases.
The first line of each test case contains one integer n (1≤n≤100) — the length of array a.
The second line of each test case contains n integers a1,a2,…,an (1≤ai≤100).

Output

For each test case print the shuffled version of the array a which is good.

Example

input

3
1
7
4
1 1 3 5
6
3 2 1 5 6 4

output

7
1 5 1 3
2 4 6 1 3 5

题意：给一组数组，要求排序，使得任意i，j，i-a[i]!=j-a[j]。

思路：水题，从大到小sort一下即可。

#include 
using namespace std;
#define pi acos(-1)
#define N 1000000007
#define ll long long
#define ull unsigned long long
#define mem(a) memset(a,0,sizeof(a))

bool cmp(int x, int y)
{
    return x > y;
}

int main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    int t;
    cin >> t;
    while(t--){
        int n;
        cin >> n;
        int s[110];
        for(int i = 1; i <= n; i++) cin >> s[i];
        if(n==1){
            cout << s[1] << endl;
        }else{
            sort(s+1,s+n+1,cmp);
            for(int i = 1; i <= n; i++) cout << s[i] << " ";
            cout << endl;
        }
    }
    return 0;
}

C. Adding Powers

Suppose you are performing the following algorithm. There is an array v1,v2,…,vn filled with zeroes at start. The following operation is applied to the array several times — at i-th step (0-indexed) you can:
either choose position pos (1≤pos≤n) and increase vpos by ki;
or not choose any position and skip this step.
You can choose how the algorithm would behave on each step and when to stop it. The question is: can you make array v equal to the given array a (vj=aj for each j) after some step?

Input

The first line contains one integer T (1≤T≤1000) — the number of test cases. Next 2T lines contain test cases — two lines per test case.
The first line of each test case contains two integers n and k (1≤n≤30, 2≤k≤100) — the size of arrays v and a and value k used in the algorithm.
The second line contains n integers a1,a2,…,an (0≤ai≤1016) — the array you’d like to achieve.

Output

For each test case print YES (case insensitive) if you can achieve the array a after some step or NO (case insensitive) otherwise.

Example

input

5
4 100
0 0 0 0
1 2
1
3 4
1 4 1
3 2
0 1 3
3 9
0 59049 810

output

YES
YES
NO
NO
YES

题意：给定一个n个元素的数组，要求判断是否能由一个全为0的数组，加上k的i次（i==1，2，3，4…)相加得到，并且i不能重复。

思路：先排序，找到最大的元素，得出i的最大值，然后把每个元素按照依次减去比自己大的k的i次，判断能不能由k的i次组成，如果能，统计i的数量，不能直接输出NO，最后判断每个i的个数，有大于1的则输出NO，反之YES。

#include 
using namespace std;
#define pi acos(-1)
#define N 1000000007
#define ll long long
#define ull unsigned long long
#define mem(a) memset(a,0,sizeof(a))

int main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    int t;
    cin >> t;
    while(t--){
        ull n, k;
        cin >> n >> k;
        ull s[100];
        for(int i = 0; i < n; i++) cin >> s[i];
        sort(s,s+n);
        int flag = 1;
        ull cnt = 0;
        ull maxn = s[n-1];
        ull x = 1;
        while(x<=maxn){
            cnt++;
            x = pow(k,cnt);
        }
        ull q[100];
        mem(q);
        for(int i = 0; i < n; i++){
            if(s[i]!=0){
                ull y = s[i];
                for(int j = cnt-1; j >= 0; j--){
                    ull z = pow(k,j);
                    if(y>=z) y -= z;
                }
                if(y==0){
                    y = s[i];
                    for(int j = cnt-1; j >= 0; j--){
                        ull z = pow(k,j);
                        if(y>=z){
                            y -= z;
                            q[j]++;
                        }
                    }
                }else{
                    flag = 0;
                    break;
                }
            }
        }
        for(int i = 0; i < cnt; i++){
            if(q[i]>1){
                flag = 0;
                break;
            }
        }
        if(flag==1){
            cout << "YES" << endl;
        }else{
            cout << "NO" << endl;
        }
    }
    return 0;
}