Codeforces Round #632 (Div. 2) D. Challenges in school №41 (模擬&思維）

Codeforces Round #632 (Div. 2) D. Challenges in school №41 (模擬&思維）

題目傳送門

題意：給定序列求使得轉身k輪剛好序列中不再存在轉身的序列的解決方案。

思路：

AC代碼：

#include<bits/stdc++.h>
using namespace std;
vector<vector<int> >ans;
int a[3050];
char c;
int main(){
	int n,k,sum=0;
	cin>>n>>k;
	for(int i=1;i<=n;i++){
		cin>>c;
		a[i]=(c=='R');//轉換爲01串,左爲0,右爲1. 
	}
	while(1){
		vector<int>now;
		for(int i=1;i<n;i++)
			if(a[i]&&!a[i+1]) now.push_back(i),swap(a[i],a[i+1]),i++;
		if(now.empty()) break;
		sum+=now.size();
		ans.push_back(now);
	}
	if(ans.size()>k||sum<k) puts("-1");
	else {
		int re=k-ans.size();//re(remain)需要填充的輪次 
		for(auto now:ans){
			while(now.size()>1&&re>0){
				printf("1 %d\n",now.back());
				now.pop_back();
				re--;
			}
			printf("%d ",now.size());
			for(auto i:now) printf("%d ",i);
			puts("");
		}
	}
	return 0;
}