手贱去开了abc,这么无聊。直接arc啊
C - 4-adjacent
Time limit : 2sec / Memory limit : 256MB
Score : 400 points
Problem Statement
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
.
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
给你一个序列,通过你的安排让这个序列满足相邻两个数的乘积是4的倍数,如果有奇数的话,而且4的倍数个数+1还比奇数多,没有2的倍数,比如1 4 1就是成立的,
2的倍数很多,这些放哪都行,只要奇数和4的倍数能凑就行
#include <bits/stdc++.h> using namespace std; const int N=1e5+5; int a[N]; int main() { int n; cin>>n; int cnt4=0,cnt2=0; for(int i=0; i<n; i++) { cin>>a[i]; a[i]%=4; if(a[i]==0) cnt4++; else if(a[i]==2) cnt2++; } int cnt=n-cnt2-cnt4; if(cnt2) { if(cnt4>=cnt){ cout<<"Yes"; return 0; } } else { if(cnt4&&(cnt4+1>=cnt)){ cout<<"Yes"; return 0; } } cout<<"No"; return 0; }
D - Grid Coloring
Time limit : 2sec / Memory limit : 256MB
Score : 400 points
Problem Statement
We have a grid with H rows and W columns of squares. Snuke is painting these squares in colors 1, 2, …, N. Here, the following conditions should be satisfied:
- For each i (1≤i≤N), there are exactly ai squares painted in Color i. Here, a1+a2+…+aN=HW.
- For each i (1≤i≤N), the squares painted in Color i are 4-connected. That is, every square painted in Color i can be reached from every square painted in Color i by repeatedly traveling to a horizontally or vertically adjacent square painted in Color i.
Find a way to paint the squares so that the conditions are satisfied. It can be shown that a solution always exists.
Constraints
- 1≤H,W≤100
- 1≤N≤HW
- ai≥1
- a1+a2+…+aN=HW
Input
Input is given from Standard Input in the following format:
H W N a1 a2 … aN
Output
Print one way to paint the squares that satisfies the conditions. Output in the following format:
c11 … c1W : cH1 … cHW
Here, cij is the color of the square at the i-th row from the top and j-th column from the left.
Sample Input 1
2 2 3 2 1 1
Sample Output 1
1 1 2 3
Below is an example of an invalid solution:
1 2 3 1
This is because the squares painted in Color 1 are not 4-connected.
Sample Input 2
3 5 5 1 2 3 4 5
Sample Output 2
1 4 4 4 3 2 5 4 5 3 2 5 5 5 3
Sample Input 3
1 1 1 1
Sample Output 3
1
#include <bits/stdc++.h> using namespace std; vector<int>a[1005]; int main() { int h,w; cin>>h>>w; int n; cin>>n; int f=0; for(int i=0; i<n; i++) { int x; scanf("%d",&x); for(int j=0;j<x;j++) a[f/w].push_back(i+1),f++; } for(int i=0; i<h; i++) { if(i&1)reverse(a[i].begin(),a[i].end()); printf("%d",a[i][0]); for(int j=1; j<w; j++) printf(" %d",a[i][j]); printf(" "); } return 0; }
E - Young Maids
Time limit : 2sec / Memory limit : 256MB
Score : 800 points
Problem Statement
Let N be a positive even number.
We have a permutation of (1,2,…,N), p=(p1,p2,…,pN). Snuke is constructing another permutation of (1,2,…,N), q, following the procedure below.
First, let q be an empty sequence. Then, perform the following operation until p becomes empty:
- Select two adjacent elements in p, and call them x and y in order. Remove x and y from p (reducing the length of p by 2), and insert x and y, preserving the original order, at the beginning of q.
When p becomes empty, q will be a permutation of (1,2,…,N).
Find the lexicographically smallest permutation that can be obtained as q.
Constraints
- N is an even number.
- 2≤N≤2×105
- p is a permutation of (1,2,…,N).
Input
Input is given from Standard Input in the following format:
N p1 p2 … pN
Output
Print the lexicographically smallest permutation, with spaces in between.
Sample Input 1
4 3 2 4 1
Sample Output 1
3 1 2 4
The solution above is obtained as follows:
p | q |
---|---|
(3,2,4,1) | () |
↓ | ↓ |
(3,1) | (2,4) |
↓ | ↓ |
() | (3,1,2,4) |
Sample Input 2
2 1 2
Sample Output 2
1 2
Sample Input 3
8 4 6 3 2 8 5 7 1
Sample Output 3
3 1 2 7 4 6 8 5
The solution above is obtained as follows:
p | q |
---|---|
(4,6,3,2,8,5,7,1) | () |
↓ | ↓ |
(4,6,3,2,7,1) | (8,5) |
↓ | ↓ |
(3,2,7,1) | (4,6,8,5) |
↓ | ↓ |
(3,1) | (2,7,4,6,8,5) |
↓ | ↓ |
() |
(3,1,2,7,4,6,8,5) |
每次取两个数,然后把这两个相邻数放到q里,使生成的q字典序最小