A. Toy CarsLittle 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?
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 Aij = 1, then Aji = 2, if Aij = 3, then Aji = 3, and if Aij = 0, then Aji = 0.
Print the number of good cars and in the next line print their space-separated indices in the increasing order.
之前傻逼一直用了个count来算0和good car的关系。明明就是car数组的个数。哎,好久没写了。
#include<iostream>
#define MAX 100
using namespace std;
int a[MAX][MAX];
int main(){
int n;
int t=0;
int car[MAX];
bool flag;
cin>>n;
for(int i=0;i<n;i++){
flag=true;
for(int j=0;j<n;j++){
cin>>a[i][j];
if(a[i][j]==1||a[i][j]==3)flag=false;
}
if(flag==true)car[t]=i+1,t++;
}
cout<<t<<endl;
for(int i=0;i<t;i++)
cout<<car[i]<<" ";
return 0;
}
本博客介绍了一个玩具车碰撞竞赛的问题,通过输入的矩阵来确定哪些玩具车是‘好车’,即在碰撞中翻转的车。通过算法找出所有翻转的车并按顺序输出其编号。
460

被折叠的 条评论
为什么被折叠?



