Problem Description

We have a sequence of length N, a=(a1,a2,…,aN). Each ai is a positive integer.

Snuke's objective is to permute the element in a so that the following condition is satisfied:

For each 1≤i≤N?1, the product of ai and ai+1 is a multiple of 4.
Determine whether Snuke can achieve his objective.

Constraints

• 2≤N≤105
• ai?is an integer.
• 1≤ai≤109

Input

Input is given from Standard Input in the following format:

N
a1 a2 … aN

Output

If Snuke can achieve his objective, print Yes; otherwise, print No.

Example

Sample Input 1

3
1 10 100

Sample Output 1

Yes
One solution is (1,100,10).

Sample Input 2

4
1 2 3 4

Sample Output 2

No
It is impossible to permute a so that the condition is satisfied.

Sample Input 3

3
1 4 1

Sample Output 3

Yes
The condition is already satisfied initially.

Sample Input 4

2
1 1

Sample Output 4

No

Sample Input 5

6
2 7 1 8 2 8

Sample Output 5

Yes

題意：給出 n 個數(shù)，問經(jīng)過任意次交換后，使得每個數(shù)乘以他的左邊是 4 的倍數(shù)，乘以右邊也是 4 的倍數(shù)，若能成功輸出 Yes，若不能輸出 No

思路：

由于每個數(shù)乘以相鄰的數(shù)是 4 的倍數(shù)，那么說明其中一個數(shù)是 4 的倍數(shù)，或者兩個數(shù)都是 2 的倍數(shù)

因此，直接統(tǒng)計這 n 個數(shù)中為 2、4 的倍數(shù)的數(shù)的個數(shù)，考慮每個位置插 2 插?4 的關(guān)系即可

Source Program

#include<iostream> #include<cstdio> #include<cstdlib> #include<string> #include<cstring> #include<cmath> #include<ctime> #include<algorithm> #include<utility> #include<stack> #include<queue> #include<vector> #include<set> #include<map> #include<bitset> #define EPS 1e-9 #define PI acos(-1.0) #define INF 0x3f3f3f3f #define LL long long const int MOD = 1E9+7; const int N = 200000+5; const int dx[] = {-1,1,0,0,-1,-1,1,1}; const int dy[] = {0,0,-1,1,-1,1,-1,1}; using namespace std; int a[N]; int main() {int n;scanf("%d",&n);int num4=0,num2=0;for(int i=1;i<=n;i++){scanf("%d",&a[i]);if(a[i]%4==0)num4++;else if(a[i]%2==0)num2++;}if(n>2*num4+num2+1)printf("No\n");else if(2*num4+num2+1==n&&num2!=0)printf("No\n");elseprintf("Yes\n");return 0; }

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