CodeForces 545A_codeforces 545div2-优快云博客

Little Susie, thanks to her older brother, likes to play with cars. Today she decided to set up a tournament between them. The process of a tournament is described in the next paragraph.

There are n toy cars. Each pair collides. The result of a collision can be one of the following: no car turned over, one car turned over, both cars turned over. A car is good if it turned over in no collision. The results of the collisions are determined by an n × n matrix А: there is a number on the intersection of the і-th row and j-th column that describes the result of the collision of the і-th and the j-th car:

- 1: if this pair of cars never collided. - 1 occurs only on the main diagonal of the matrix.
0: if no car turned over during the collision.
1: if only the i-th car turned over during the collision.
2: if only the j-th car turned over during the collision.
3: if both cars turned over during the collision.

Susie wants to find all the good cars. She quickly determined which cars are good. Can you cope with the task?

Input

The first line contains integer n (1 ≤ n ≤ 100) — the number of cars.

Each of the next n lines contains n space-separated integers that determine matrix A.

It is guaranteed that on the main diagonal there are - 1, and - 1 doesn't appear anywhere else in the matrix.

It is guaranteed that the input is correct, that is, if A_ij = 1, then A_ji = 2, if A_ij = 3, then A_ji = 3, and if A_ij = 0, then A_ji = 0.

Output

Print the number of good cars and in the next line print their space-separated indices in the increasing order.

Sample Input

Input

3-1 0 00 -1 10 2 -1

Output

21 3

Input

4-1 3 3 33 -1 3 33 3 -1 33 3 3 -1

Output

///这是本次这周赛中传说最简单的一道题，抠英文抠了好半天就是读不懂题意，英语是硬伤啊>_<最后看明白的差不多了，输入的n是指小汽车的数量，矩阵中的-1是指小车不会和自己相撞，0表示两个车都不会撞坏，1表示矩阵中该行所表示的车号被撞坏，2表示矩阵中该列所表示的车号被撞坏，3表示都被撞坏。以下有两种方法：

法一：

#include<stdio.h>
#include<string.h>
int s[110][110], a[110], b[110], vis[110];

int main()
{
    int n, i, j;
    scanf("%d", &n);
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<n;j++)
        {
            scanf("%d", &s[i][j]);
        }
    }
    memset(vis,0,sizeof(vis));
    memset(a,0,sizeof(a));
    for(int i=0;i<n;i++)
    {
        a[i]=i;
    }
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<n;j++)
        {
            if(s[i][j]==1)
            {
                vis[i]=1;
            }
            else if(s[i][j]==2)
            {
                vis[j]=1;
            }
            else if(s[i][j]==3)
            {
                vis[i]=vis[j]=1;
            }
        }
    }
    int t=0, sum=0;
    for(int i=0;i<n;i++)
    {
        if(vis[i]==0)
        {
            b[t++]=a[i];
            sum++;
        }
    }
    printf("%d\n", sum);
    if(sum!=0)
    {
        for(int i=0;i<t;i++)
        {
            printf("%d", b[i]+1);
            if(i<t-1)
                printf(" ");
        }
    }
    return 0;
}

法二：

#include<stdio.h>
int main()
{
    int n;
    int s[110][110];
    scanf("%d", &n);
    for(int i=0; i<n; i++)
    {
        for(int j=0; j<n; j++)
        {
            scanf("%d", &s[i][j]);
        }
    }
    int f;
    int a[110], t=0, sum=0;
    for(int i=0; i<n; i++)
    {
        f=1;
        for(int j=0; j<n; j++)
        {
            if(s[i][j]==1 || s[i][j]==3)
            {
                f=0;
                break;
            }
        }
        if(f==1)
        {

a[t++]=i+1;//

            sum++;
        }
    }
    printf("%d\n", sum);
    if(sum!=0)
    {
        for(int i=0; i<t; i++)
        {
            printf("%d", a[i]);
            if(i<t-1)
                printf(" ");
        }
    }
    return 0;
}