E - Dima and a Bad XOR CodeForces - 1151B

最新推荐文章于 2023-01-06 13:59:45 发布

DeepGoAI

最新推荐文章于 2023-01-06 13:59:45 发布

阅读量381

点赞数

CC 4.0 BY-SA版权

分类专栏： # ACM # 异或

本文链接：https://blog.youkuaiyun.com/Willen_/article/details/90725885

ACM 同时被 2 个专栏收录

56 篇文章

订阅专栏

异或

1 篇文章

订阅专栏

本文详细解析了 CodeForces-1151B 的解题思路，介绍了如何从一个 n*m 的矩阵中选取一行的一个元素，使所有选中元素的异或结果大于零。通过实例演示了算法的实现过程。

E - Dima and a Bad XOR

CodeForces - 1151B

Student Dima from Kremland has a matrix aa of size n×mn×m filled with non-negative integers.

He wants to select exactly one integer from each row of the matrix so that the bitwise exclusive OR of the selected integers is strictly greater than zero. Help him!

Formally, he wants to choose an integers sequence c1,c2,…,cnc1,c2,…,cn (1≤cj≤m1≤cj≤m) so that the inequality a1,c1⊕a2,c2⊕…⊕an,cn>0a1,c1⊕a2,c2⊕…⊕an,cn>0 holds, where ai,jai,j is the matrix element from the ii-th row and the jj-th column.

Here x⊕yx⊕y denotes the bitwise XOR operation of integers xx and yy.

Input

The first line contains two integers nn and mm (1≤n,m≤5001≤n,m≤500) — the number of rows and the number of columns in the matrix aa.

Each of the next nn lines contains mm integers: the jj-th integer in the ii-th line is the jj-th element of the ii-th row of the matrix aa, i.e. ai,jai,j (0≤ai,j≤10230≤ai,j≤1023).

Output

If there is no way to choose one integer from each row so that their bitwise exclusive OR is strictly greater than zero, print "NIE".

Otherwise print "TAK" in the first line, in the next line print nn integers c1,c2,…cnc1,c2,…cn (1≤cj≤m1≤cj≤m), so that the inequality a1,c1⊕a2,c2⊕…⊕an,cn>0a1,c1⊕a2,c2⊕…⊕an,cn>0 holds.

If there is more than one possible answer, you may output any.

Examples

Input

Output

NIE

Input

2 3
7 7 7
7 7 10

Output

TAK
1 3

Note

In the first example, all the numbers in the matrix are 00, so it is impossible to select one number in each row of the table so that their bitwise exclusive OR is strictly greater than zero.

In the second example, the selected numbers are 77 (the first number in the first line) and 1010 (the third number in the second line), 7⊕10=137⊕10=13, 1313 is more than 00, so the answer is found.

题目大意：

给定一个n,m

再给出一个n*m的矩阵

然后让你在每一行中取走一个数，组成的这n个数进行异或能够得到一个大于0的数

思路:

其实我们可以先取一组数据，看看是否异或大于0，如果是的话，直接输出，如果不是的话，再从每一行找出与之前这个数不一样大小的数字（因为不相等的数字异或为0，这个很关键），如果有，那么就存在，最后遍历完毕，仍然没有，那就是真的没有

#include<iostream>
#include<cstdio>
using namespace std;

const int MAXN = 510;

int map[MAXN][MAXN] ={0};
int N,M;
//int out[MAXN] = 0;
int main(){
	
	scanf("%d%d",&N,&M);
	
	int ans = 0;
	for(int i=1;i<=N;i++){
		for(int j=1;j<=M;j++){
			scanf("%d",&map[i][j]);
		}	
		ans^=map[i][1];
	}
	
	if(ans>0){
		printf("TAK\n");
		for(int i=1;i<=N;i++){
			if (i==1)printf("1");
			else printf(" 1");
		}
				
		return 0;
	}else{
		for(int i=1;i<=N;i++){
			for(int j=2;j<=M;j++){
				int ii,ij;
				if(map[i][j]!=map[i][1]){
					ii = i;
					ij = j;
					printf("TAK\n");
					for(int i=1;i<=N;i++){
						if(i==ii)printf("%d",ij);
						else printf("1");
						if(i<N) printf(" ");
					}
					
					printf("\n");
					return 0;
				}
			}
		}
	}
		
	printf("NIE\n");

	return 0;
}