周赛二 CodeForces 545A

Description

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

0

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#include<iostream>#include<cstdio>#include<cstring>using namespace std;int main(){    int n;    int b[105],a[105][106];    while(cin>>n)    {        for(int i=0; i<n; i++)            b[i]=i+1;        for(int i=0; i<n; i++)            for(int j=0; j<n; j++)            {                cin>>a[i][j];                if(a[i][j]==1)                    b[i]=-1;                else if(a[i][j]==2)                    b[j]=-1;                else if(a[i][j]==3)                {                    b[i]=-1;                    b[j]=-1;                }            }        int num=0;        for(int i=0; i<n; i++)            if(b[i]!=-1)                num++;        cout<<num<<endl;        int kk=0;        for(int i=0; i<n; i++)            if(b[i]==i+1)            {                if(kk==0)                    kk=1;                else                    cout<<" ";                cout<<i+1;                if(i==n-1)cout<<endl;            }    }}

两种最显然的方法。
1.用个数组标记一下每辆车是否被撞，最后遍历一下。
2.分别对每辆车去遍历一遍跟它有关的行和列，判断一下是否被撞过即可。