D. Challenges in school №41 模拟
解题思路：判断形成目标状态所须的最小，最大时间，与标准时间对比，在其区间内，则可行，然后模拟
1 #include<bits/sdtc++.h> 2 using namespace std; 3 #define ll long long 4 int n, k; 5 ll maxk = 0ll, mink = 0ll, d; 6 char str[3050]; 7 vector<int> v[300050]; 8 void solve() 9 { 10 int p = 0, p2 = 0, kk = mink; 11 while(kk<k){ 12 cout << "1 " << v[p][p2] << endl; 13 p2++; 14 if(p2==v[p].size()){ 15 p2=0; 16 p++; 17 } 18 else kk++; 19 } 20 cout << v[p].size() - p2; 21 for(int i = p2; i < v[p].size(); i++) 22 cout << " " << v[p][i]; 23 cout << endl; 24 while(p<mink){ 25 cout << v[p].size(); 26 for(int i = 0; i < v[p].size(); i++) 27 cout << " " << v[p][i]; 28 cout << endl; 29 p++; 30 } 31 } 32 void check(int p){ 33 for(int i = 1; i < n; ){ 34 if(str[i] == 'R' && str[i+1] == 'L'){ 35 str[i] = 'L'; 36 str[i+1] = 'R'; 37 v[p].push_back(i); 38 i += 2; 39 } 40 else i++; 41 } 42 } 43 int main() 44 { 45 ios::sync_with_stdio(false); 46 cin.tie(0); cout.tie(0); 47 cin >> n >> k >> (str+1); 48 check(0); 49 while(v[mink].size()&&mink<=k){ 50 maxk += v[mink].size(); 51 mink++; 52 check(mink); 53 } 54 if(mink<=k&&k<=maxk) solve(); 55 else cout << -1 << endl; 56 }
