Codeforces Round #303 (Div.2)-A. Toy Cars(模拟)

A. Toy Carstime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputLittle 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 turnedover, 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-thcolumn that describes the result of the collision of the і-th and the j-thcar: - 1: if this pair of cars never collided.  - 1 occursonly on the main diagonal of the matrix.0: if no car turned over during the collision.1: if only the i-thcar turned over during the collision.2: if only the j-thcar 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?InputThe first line contains integer n (1 ≤ n ≤ 100)— the number of cars.Each of the next n lines contains n space-separatedintegers that determine matrix A.It is guaranteed that on the main diagonal there are  - 1, and  - 1 doesn'tappear anywhere else in the matrix.It is guaranteed that the input is correct, that is, if Aij = 1,then Aji = 2,if Aij = 3,then Aji = 3,and if Aij = 0,then Aji = 0.OutputPrint the number of good cars and in the next line print their space-separated indices in the increasing order.Examplesinput

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：

有1,2,3三个操作，1是i车翻了，2是j车翻了，3是i，j两辆车都翻了，问你还有几辆车没翻。

AC代码：

#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>#include<vector>using namespace std;int main(){#ifdef zscfreopen("input.txt","r",stdin);#endifint n,m,i,j,k;while(~scanf("%d",&n)){int vis[110]={0};for(i=1;i<=n;++i){for(j=1;j<=n;++j){scanf("%d",&k);if(j>i){if(k==1)vis[i]=true;if(k==2)vis[j]=true;if(k==3)vis[i]=vis[j]=true;}}}vector<int> ve;for(i=1;i<=n;++i){if(!vis[i])ve.push_back(i);}printf("%d\n",ve.size());for(i=0;i<ve.size();++i){printf("%d ",ve[i]);}printf("\n");}return 0;}