A. Toy Cars

A. Toy Cars

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 test(s)

input

3
-1 0 0
0 -1 1
0 2 -1

output

2
1 3 

input

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

output

0

题意：n表示小车的数量，-1表示小车自己不会与自己相撞，1表示该行所表示的车号被撞坏，2表示该列所表示的车号被撞坏，3表示都被撞坏

法1：

#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2

#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;
}