Codeforces1016D题解

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一道好玩的构造题

Description

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。

$$n, m \leq 100$$

Samples

Input #1

1
2
3
2 3
2 9
5 3 13

Output #1

1
2
3
YES
3 4 5
6 7 8

Input #2

1
2
3
3 3
1 7 6
2 15 12

Output #2

1
NO

Solution

一道需要脑洞开大的好玩构造题~

考虑异或的性质$a \text{ } \rm{xor} \text{ } 0 = a$, 即一个数异或0之后的结果仍然是其本身.

那么我们把目标矩阵除了最后一列和最后一行, 都直接填上0.

这样的后果是, 除了右下角的元素, 其它的元素直接填上对应行/列的异或和即可.

右下角的元素可以通过异或的另一个性质$a \text{ } \rm{xor} \text{ } b = c \implies b = c \text{ } \rm{xor} \text{ } a$推出来.

Code

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/* Headers */
#include<algorithm>
#include<iostream>
#include<cstring>
#include<climits>
#include<cstdio>
#include<cctype>
#include<vector>
#include<cmath>
#include<array>
#include<queue>
#include<stack>
#include<map>
#define FOR(i,a,b,c) for(int i=(a);i<=(b);i+=(c))
#define ROF(i,a,b,c) for(int i=(a);i>=(b);i-=(c))
#define FORL(i,a,b,c) for(long long i=(a);i<=(b);i+=(c))
#define ROFL(i,a,b,c) for(long long i=(a);i>=(b);i-=(c))
#define FORR(i,a,b,c) for(register int i=(a);i<=(b);i+=(c))
#define ROFR(i,a,b,c) for(register int i=(a);i>=(b);i-=(c))
#define RevEdge(x) x^1
#define lowbit(x) x&(-x)
#define LeftChild(x) x<<1
#define RightChild(x) (x<<1)+1
#define CLOSE_IN() fclose(stdin);
#define CLOSE_OUT() fclose(stdout);
#define FILE_IN(x) freopen(x,"r",stdin);
#define FILE_OUT(x) freopen(x,"w",stdout);
#define IOS(x) std::ios::sync_with_stdio(x)
#define Dividing() printf("-----------------------------------\n");
namespace FastIO{
    const int BUFSIZE = 1 << 20;
    char ibuf[BUFSIZE],*is = ibuf,*its = ibuf;
    char obuf[BUFSIZE],*os = obuf,*ot = obuf + BUFSIZE;
    inline char getch(){
        if(is == its)
            its = (is = ibuf)+fread(ibuf,1,BUFSIZE,stdin);
        return (is == its)?EOF:*is++;
    }
    inline int getint(){
        int res = 0,neg = 0,ch = getch();
        while(!(isdigit(ch) || ch == '-') && ch != EOF)
            ch = getch();
        if(ch == '-'){
            neg = 1;ch = getch();
        }
        while(isdigit(ch)){
            res = (res << 3) + (res << 1)+ (ch - '0');
            ch = getch();
        }
        return neg?-res:res;
    }
    inline void flush(){
        fwrite(obuf,1,os - obuf,stdout);
        os = obuf;
    }
    inline void putch(char ch){
        *os++ = ch;
        if(os == ot)	flush();
    }
    inline void putint(int res){
        static char q[10];
        if(res == 0) putch('0');
        else if(res < 0){putch('-');res = -res;}
        int top = 0;
        while(res){
            q[top++] = res % 10 + '0';
            res /= 10;
        }
        while(top--)	putch(q[top]);
    }
    inline void space(bool x){
    	if(!x) putch('\n');
    	else putch(' ');
    }
}
inline void read(int &x){
    int rt = FastIO::getint();
    x = rt;
}
inline void print(int x,bool enter){
    FastIO::putint(x);
    FastIO::flush();
    FastIO::space(enter);
}
/* definitions */
constexpr int MAXN = 1e2 + 10;
int n, m, ans;
std::array<int, MAXN> Iline{0}, Jline{0}; 
/* functions */
int main(int argc,char *argv[]){
	scanf("%d %d", &n, &m);
	FOR(i, 1, n, 1) scanf("%d", &Iline[i]), ans ^= Iline[i];
	FOR(i, 1, m, 1) scanf("%d", &Jline[i]), ans ^= Jline[i];
	if(ans != 0) return puts("NO") & 0;
	printf("YES\n"); ans = Jline[m];
	FOR(i, 1, n - 1, 1) {
		FOR(j, 1, m - 1, 1) printf("0 ");
		printf("%d\n", Iline[i]);
		ans = ans ^ Iline[i];
	}
	FOR(i, 1, m - 1, 1) printf("%d ", Jline[i]);
	printf("%d\n", ans);
	return 0;
}

THE END