题目描述
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 n n rows and m m m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1,a2,...,an a_{1},a_{2},...,a_{n} a1,a2,...,an denotes the xor of elements in rows with indices 1 1 1 , 2 2 2 , ..., n n n , respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1,b2,...,bm b_{1},b_{2},...,b_{m} b1,b2,...,bm denotes the xor of elements in columns with indices 1 1 1 , 2 2 2 , ..., m m m , respectively.
Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.
翻译:已知一个n×m的矩阵,每行每列元素的异或和,请构造一个满足要求的矩阵。若不存在,输出"NO",否则输出"YES"和矩阵。
输入输出格式
输入格式:The first line contains two numbers n n n and $ m (2<=n,m<=100) $ — the dimensions of the matrix.
The second line contains n n n numbers $ a_{1},a_{2},...,a_{n} (0<=a_{i}<=10^{9}) $ , where ai a_{i} ai is the xor of all elements in row i i i .
The third line contains m m m numbers $ b_{1},b_{2},...,b_{m} (0<=b_{i}<=10^{9}) $ , where bi b_{i} bi is the xor of all elements in column i i i .
输出格式: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 n n rows of m m m numbers in each $ c_{i1},c_{i2},... ,c_{im} (0<=c_{ij}<=2·10^{9}) $ — the description of the matrix.
If there are several suitable matrices, it is allowed to print any of them.
输入输出样例
2 3 2 9 5 3 13
YES 3 4 5 6 7 8
3 3 1 7 6 2 15 12
题目解析
Code
#include<iostream> #include<cstdio> using namespace std; const int MAXN = 100 + 5; int n,m; int a[MAXN],b[MAXN]; int ans1,ans2; int main() { scanf("%d%d",&n,&m); for(int i = 1;i <= n;i++) { scanf("%d",&a[i]); ans1 ^= a[i]; } for(int i = 1;i <= m;i++) { scanf("%d",&b[i]); ans2 ^= b[i]; } if(ans1 ^ ans2) { printf("NO "); return 0; } else printf("YES "); for(int i = 1;i < n;i++) { for(int j = 1;j < m;j++) printf("0 "); printf("%d ",a[i]); } for(int i = 1;i < m;i++) { printf("%d ",b[i]); } int ans = b[m]; for(int i = 1;i < n;i++) ans ^= a[i]; printf("%d",ans); return 0; }