D. Vasya And The Matrix Educational Codeforces Round 48 (Rated for Div. 2)
D. Vasya And The Matrix
Educational Codeforces Round 48 (Rated for Div. 2)
D. Vasya And The Matrix
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Now Vasya is taking an exam in mathematics. In order to get a good mark, Vasya needs to guess the matrix that the teacher has constructed!
Vasya knows that the matrix consists of n rows and m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1, a2, ..., an denotes the xor of elements in rows with indices 1, 2, ..., n, respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1, b2, ..., bm denotes the xor of elements in columns with indices 1, 2, ..., m, respectively.
Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.
Input
The first line contains two numbers n and m (2 ≤ n, m ≤ 100) — the dimensions of the matrix.
The second line contains n numbers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the xor of all elements in row i.
The third line contains m numbers b1, b2, ..., bm (0 ≤ bi ≤ 109), where bi is the xor of all elements in column i.
Output
If there is no matrix satisfying the given constraints in the first line, output "NO".
Otherwise, on the first line output "YES", and then n rows of m numbers in each ci1, ci2, ... , cim (0 ≤ cij ≤ 2·109) — the description of the matrix.
If there are several suitable matrices, it is allowed to print any of them.
Examples
input
2 3 2 9 5 3 13
output
YES 3 4 5 6 7 8
input
3 3 1 7 6 2 15 12
output
NO
题意：给你一个n*m的矩阵的每一行的数异或的值和每一列的异或值，问是否可以构造出这个矩阵来，如果不能，则输出NO，可以则输出YESh和构造的这个矩阵。
思路：一道思维题，首先我们可以想到把所有的行和所有的列异或值再全部异或，得到的值如果不为0，则必定不存在这样的矩阵，反之，则一定能构造出来这样的矩阵。其实只要构造第一行和第一列就可以了（也可以最后一行和最后一列），将第一行的第二个开始的值全部构造成他们的列异或值，第一列的第二个开始的值全部构造成它们的列的异或值，除了第一行第一列的那个值之外，剩下的全部构造成0，那么我们主要要构造的就是c[0][0],可以构造成c[0][0]=b[0]^a[1]^a[2]^....a[n-1],在第一列的方向上，很明显，他们的异或值为b[0]，在第一行的方向上，异或值为(b[0]^a[1]^a[2]^.....a[n-1])^b[1]^b[2]^....^b[m-1]==a[0],因为他们总的异或值为0，这样就构造完毕了。
#include
using namespace std;
#define ll long long
#define lowbit(x) (x&-x)
const int maxn=int(1e2)+10;
const int inf=0x3f3f3f3f;
const ll mod=998244353;
using namespace std;
int a[maxn],b[maxn];
int c[maxn][maxn];
int n,m;
void debug() {
for(int i=0; i