CS Academy Gcd Rebuild

题目链接：https://csacademy.com/contest/archive/task/gcd-rebuild/statement/

题目大意：给出一个N*M的矩阵，其中第i行j列表示gcd(a[i], b[j])，现在不知道数组a，b，给出这个矩阵，求a，b。如果多组解，输出其中一组，若无解，输出-1.

解题思路：观察观察（真的是观察出来的），发现如果有解，那么可以按照如下方式构造解：

1 for(int i = 1; i <= n; i++){
2     ll tmp = 1, la;
3     for(int j = 1; j <= m; j++){
4         la = tmp;
5         tmp *= c[i][j];
6         tmp /= gcd(la, c[i][j]);
7     }
8     a[i] = tmp;
9 }

对于每一列也是同样的方式。构造之后判断一下是否符合条件即可。

代码：

 1 const int maxn = 3e2 + 5;
 2 int n, m;
 3 ll c[maxn][maxn];
 4 ll a[maxn], b[maxn];
 5 
 6 ll gcd(ll x, ll y){
 7     return y == 0? x: gcd(y, x % y);
 8 }
 9 void solve(){
10     for(int i = 0; i <= n; i++) c[i][0] = c[0][i] = 1;
11     bool flag = true;
12     for(int i = 1; i <= n; i++){
13         ll tmp = 1, la;
14         for(int j = 1; j <= m; j++){
15             la = tmp;
16             tmp *= c[i][j];
17             tmp /= gcd(la, c[i][j]);
18         }
19         a[i] = tmp;
20         if(a[i] > 1e9) {
21             flag = false;
22             break;
23         }
24     }
25     if(!flag){
26         puts("-1");
27         return;
28     }
29     for(int i = 1; i <= m; i++){
30         ll tmp = 1, la;
31         for(int j = 1; j <= n; j++){
32             la = tmp;
33             tmp *= c[j][i];
34             tmp /= gcd(c[j][i], la);
35         }
36         b[i] = tmp;
37         if(b[i] > 1e9) {
38             flag = false;
39             break;
40         }
41     }
42     if(!flag){
43         puts("-1");
44         return;
45     }
46     for(int i = 1; i <= n; i++){
47         for(int j = 1; j <= m; j++){
48             if(c[i][j] != gcd(a[i], b[j])){
49                 flag = false;
50                 break;
51             }
52         }
53         if(!flag) break;
54     }
55     if(!flag){
56         puts("-1");
57         return;
58     }
59     for(int i = 1; i <= n; i++) printf("%lld ", a[i]);
60     puts("");
61     for(int i = 1; i <= m; i++) printf("%lld ", b[i]);
62 }
63 int main(){
64     scanf("%d %d", &n, &m);
65     for(int i = 1; i <= n; i++){
66         for(int j = 1; j <= m; j++){
67             scanf("%lld", &c[i][j]);
68         }
69     }
70     solve();
71 }

题目：

Gcd Rebuild

Time limit: 1000 ms
Memory limit: 256 MB

 

Consider two arrays: VV of size NN and UU of size MM. The elements of both arrays are integers between 11 and 10^910​9​​.

You are given a matrix AA where A_{i,j} = \gcd(V_i, U_j)A​i,j​​=gcd(V​i​​,U​j​​) and \gcdgcd refers to the greatest common divisor. You are asked to find VV and UU.

Standard input

The first line contains two integers NN and MM.

Each of the next NN lines contains MM integers, representing the elements of AA.

Standard output

If there is no solution, or you cannot find a solution where the elements of VV and UU are in the range [1, 10^9][1,10​9​​], output -1−1.

Otherwise, print the NN elements of VV on the first line and the MM elements of UU on the second line. If the solution is not unique you can output any of them.

Constraints and notes

  • 1 \leq N, M \leq 3001≤N,M≤300 
  • 1 \leq A_{i, j} \leq 10^91≤A​i,j​​≤10​9​​ 
InputOutputExplanation
 
3 3
1 2 2
3 2 2
3 1 1

2 6 3 
3 2 2

gcd(2, 3)=1gcd(2,3)=1

gcd(2, 2)=2gcd(2,2)=2

gcd(6, 3)=3gcd(6,3)=3

gcd(6, 2)=2gcd(6,2)=2

gcd(3, 3)=3gcd(3,3)=3

gcd(3, 2)=1gcd(3,2)=1

 
3 2
2 2
2 2
4 2

2 2 4 
4 2

gcd(2, 2)=2gcd(2,2)=2

gcd(2, 4)=2gcd(2,4)=2

gcd(4, 4)=4gcd(4,4)=4

 
2 2
1 1
1 1

1 1 
1 1

gcd(1, 1)=1gcd(1,1)=1 

3 3
4 2 4
2 4 2
4 2 4

-1

There is no solution
posted @ 2017-09-10 10:52  EricJeffrey  阅读(167)  评论(0)  编辑  收藏  举报