CF 1371D Grid-00100

題目：

A mad scientist Dr.Jubal has made a competitive programming task. Try to solve it!

You are given integers n,k. Construct a grid AA with size n×n consisting of integers 0 and 1. The very important condition should be satisfied: the sum of all elements in the grid is exactly k. In other words, the number of 1 in the grid is equal to k.

Let's define:

• Ai,j as the integer in the i-th row and the j-th column.
• Ri=Ai,1+Ai,2+...+Ai,n (for all 1≤i≤n).
• Cj=A1,j+A2,j+...+An,j (for all 1≤j≤n).
• In other words, RiRi are row sums and CjCj are column sums of the grid AA.
• For the grid A let's define the value f(A)=(max(R)−min(R))^2+(max(C)−min(C))^2 (here for an integer sequence X we define max(X) as the maximum value in X and min(X) as the minimum value in X).

Find any grid A, which satisfies the following condition. Among such grids find any, for which the value f(A) is the minimum possible. Among such tables, you can find any.

 

思路：咱們容易想到讓max儘量的小，min儘量的大。經過畫圖把1合理分配，容易獲得一個畫圖的方式，每次都畫右偏的對角線便可，若是超出數組則對應左邊標記1便可。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <queue>
 5 #include <string>
 6 #include <vector>
 7 #include <cmath>
 8  
 9 using namespace std;
10  
11 #define ll long long
12 #define pb push_back
13 #define fi first
14 #define se second
15  
16 const int N = 2e5 + 10;
17 int a[310][310];
18  
19 void solve()
20 {      
21     int T;
22     cin >> T;
23     while(T--){
24         int n, k;
25         cin >> n >> k;
26  
27         for(int i = 1; i <= n; ++i){
28             for(int j = 1; j <= n; ++j)
29                 a[i][j] = 0;
30         }
31 
32         //畫圖
33         int d = 0;
34         while(1){
35             for(int i = 1; i <= n; ++i){
36                 if(k == 0) break;
37                 a[i][(i + d) == n ? n : (i + d) % n] = 1;
38                 k--;
39             }
40             if(k == 0) break;
41             d++;
42         }
43  
44         int maxr, minr, maxc, minc;
45         maxr = maxc = -1;
46         minr = minc = 1e9;
47         for(int i = 1; i <= n; ++i){
48             int cnt = 0;
49             for(int j = 1; j <= n; ++j){
50                 if(a[i][j]) cnt++;
51             }
52             maxr = max(maxr, cnt);
53             minr = min(minr, cnt);
54         }
55  
56         for(int j = 1; j <= n; ++j){
57             int cnt = 0;
58             for(int i = 1; i <= n; ++i){
59                 if(a[i][j]) cnt++;
60             }
61             maxc = max(maxc, cnt);
62             minc = min(minc, cnt);
63         }
64         //cout << "min_ans = ";
65         cout << (maxr - minr) * (maxr - minr) + (maxc - minc) * (maxc - minc) << endl;
66         for(int i = 1; i <= n; ++i){
67             for(int j = 1; j <= n; ++j){
68                 cout << a[i][j];
69             }
70             cout << endl;
71         }
72     }
73 }
74  
75 int main()
76 {
77     ios::sync_with_stdio(false);
78     cin.tie(0);
79     cout.tie(0); 
80     solve();
81  
82     return 0;
83 }