What is the difference between the snapshot 10273 Eat and Not to Eat?

 
Idea: brute-force Solution
Analysis:
1. The number of cows that have not been eaten must be the number of days that have been last eaten.
2. No other method can only be violent. For n cycles of native cattle, obtain the minimum public multiple, And then solve the problem by brute force within 2 Public multiples.

Code:

 

#include<vector>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int MAXN = 1010;int n , lcm;vector<int>v[MAXN];bool isEat[MAXN];int gcd(int a , int b){    return b == 0 ? a : gcd(b , a%b);}void init(){    memset(isEat , false , sizeof(isEat));    for(int i = 0 ; i < MAXN ; i++)        v[i].clear();}void solve(){    int index = 0;    int notEat = n;    int numDay = 0;    while(index < 2*lcm){        int min = 1<<30;        int minIndex = -1;        for(int i = 0 ; i < n ; i++){           if(!isEat[i]){               int size = v[i].size();                int tmp = v[i][index%size];                  if(min > tmp){                  min = tmp;                  minIndex = i;               }               else{                  if(min == tmp)                      minIndex = -1;               }           }          }        index++;        if(minIndex != -1){           notEat--;           numDay = index;           isEat[minIndex] = true;        }    }    printf("%d %d\n" , notEat , numDay);}int main(){    int Case , m , x;    scanf("%d" , &Case);    while(Case--){         scanf("%d" , &n);         init();         bool isFirst = true;         for(int i = 0 ; i < n ; i++){             scanf("%d" , &m);              if(isFirst){                lcm = m;                 isFirst = false;             }             lcm = lcm/gcd(lcm , m)*m;             while(m--){                  scanf("%d" , &x);                  v[i].push_back(x);             }         }         solve();    }}