Educational Codeforces Round 95 (Rated for Div. 2) A~D

直接求最少需要多少个火柴就行

#include<bits/stdc++.h>
#include<string.h>
using namespace std;
#define rep(i,j,k) for(LL i=(j); i<(k); ++i)
#define pb push_back
#define PII pair<LL,LL>
#define PLL pair<long long, long long>
#define ini(a,j) memset(a,j,sizeof a)
#define rrep(i,j,k) for(LL i=j; i>=k; --i)
#define fi first
#define se second
#define LL long long
#define beg begin()
#define ed end()
#define all(x) x.begin(),x.end()


int main(int argc, char const *argv[])
{
//	#define DEBUG
    	#ifdef DEBUG
		freopen("1.dat","r",stdin);
	#endif
	LL _;
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>_;
	
	while(_--){
		LL a,b,c;
		cin>>a>>b>>c;
		LL ans = 0;
		LL cur = 1;
		LL need = c+b*c-1;
		ans = need/(a-1) + (need%(a-1)!=0);
		cout<<ans+c<<endl;
		
	}
	return 0;
}

把可以挪动的位置的值从大到小排序

#include<bits/stdc++.h>
#include<string.h>
using namespace std;
#define rep(i,j,k) for(LL i=(j); i<(k); ++i)
#define pb push_back
#define PII pair<LL,LL>
#define PLL pair<long long, long long>
#define ini(a,j) memset(a,j,sizeof a)
#define rrep(i,j,k) for(LL i=j; i>=k; --i)
#define fi first
#define se second
#define LL long long
#define beg begin()
#define ed end()
#define all(x) x.begin(),x.end()


int main(int argc, char const *argv[])
{
//	#define DEBUG
    	#ifdef DEBUG
		freopen("1.dat","r",stdin);
	#endif
	LL _;
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>_;
	
	while(_--){
		int n;
		cin>>n;
		vector<int> a(n+1,0),b(n+1,0);
		rep(i,0,n) cin>>a[i+1];
		rep(i,0,n) cin>>b[i+1];
		
		vector<int>  temp;
		
		rep(i,0,n){
			if(b[i+1])
				continue;
			else{
				temp.push_back(a[i+1]);
			}
		}
		sort(all(temp), std::greater<int>());
		int cnt = 0;
		for(int i=1; i<=n; i++){
			if(b[i]==0)
				a[i]=temp[cnt++];
		}
		for(int i=1; i<=n; i++){
			cout<<a[i]<<" ";
		}
		cout<<endl;
	}
	return 0;
}

简单的dp问题
设dp[i][0]， dp[i][1] 分别是当第i个位置
是朋友和自己回合的回合的最少跳过次数
dp[i][0] = min(dp[i-1][1]+a[i], dp[i-1][1]+a[i]+a[i-1])
dp[i][1] = min(dp[i-1][0], dp[i-1][0])

#include<bits/stdc++.h>
#include<string.h>
using namespace std;
#define rep(i,j,k) for(LL i=(j); i<(k); ++i)
#define pb push_back
#define PII pair<LL,LL>
#define PLL pair<long long, long long>
#define ini(a,j) memset(a,j,sizeof a)
#define rrep(i,j,k) for(LL i=j; i>=k; --i)
#define fi first
#define se second
#define LL long long
#define beg begin()
#define ed end()
#define all(x) x.begin(),x.end()


int main(int argc, char const *argv[])
{
//	#define DEBUG
    	#ifdef DEBUG
		freopen("1.dat","r",stdin);
		freopen("ans.dat","w",stdout);
	#endif
	LL _;
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>_;
	
	while(_--){
		int n;
		cin>>n;
		vector<int> a(n);
		rep(i,0,n) cin>>a[i];
		vector<vector<int>> dp(n+15, vector<int>(2,10000000));
		dp[0][0]=0;  //0 is friend`s round
		dp[0][1]=0;
		for(int i=1; i<=n; i++){
			dp[i][0]=min(dp[i-1][1]+a[i-1], dp[i][0]);
			if(i>=2)
				dp[i][0]=min(dp[i-2][1]+a[i-1]+a[i-2], dp[i][0]);
			if(i>1)
				dp[i][1] = min(dp[i-1][0], dp[i][1]);
			if(i>2)
				dp[i][1] = min(dp[i-2][0], dp[i][1]);
		}
		cout<<min(dp[n][0], dp[n][1])<<endl;
	}
	return 0;
}

肯定是从两边往中间收缩
留下最大的缝隙
第一次知道迭代器还能这么玩
但是懒得写函数的写了个又臭又长的代码

#include<bits/stdc++.h>
#include<string.h>
using namespace std;
#define rep(i,j,k) for(LL i=(j); i<(k); ++i)
#define pb push_back
#define PII pair<LL,LL>
#define PLL pair<long long, long long>
#define ini(a,j) memset(a,j,sizeof a)
#define rrep(i,j,k) for(LL i=j; i>=k; --i)
#define fi first
#define se second
#define LL long long
#define beg begin()
#define ed end()
#define all(x) x.begin(),x.end()


int main(int argc, char const *argv[])
{
//	#define DEBUG
    	#ifdef DEBUG
		freopen("1.dat","r",stdin);
//		freopen("ans.dat","w",stdout);
	#endif
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int n;
	cin>>n;
	vector<int> a(n);
	int q;
	cin>>q;
	rep(i,0,n) cin>>a[i];
	int ans = 0;
	sort(all(a));
	map<int,int> res;
	set<int> temp;
	temp.insert(a[0]);
	for(int i=1; i<n; i++){
		temp.insert(a[i]);
		ans += a[i]-a[i-1];
		res[a[i]-a[i-1]]++;
	}
	int gg = ans;
	if(res.size())
		ans -= (*(--res.end())).fi;
	cout<<ans<<endl;
	int t,x;
	while(q--){
		cin>>t>>x;
		auto pos = temp.lower_bound(x);
		if(t==0){
			// remove a new 
			int num = *pos;
			if(temp.size()==1){
				cout<<gg<<endl;
				temp.erase(pos);
				continue;
			}
			if(pos==temp.begin()){
				++pos;
				res[*pos-num]--;
				if(res[*pos-num]==0)
					res.erase(res.find(*pos-num));
				gg -= *pos-num;
			}else{
				--pos;
				int num2= *pos;
				res[num-num2]--;
				gg -= num-num2;
				if(res[num-num2]==0)
					res.erase(res.find(num-num2));
				++pos;
				++pos;
				int num3;
				if(pos!=temp.end()){
					num3 = *pos;
					res[num3-num]--;
					gg -= num3-num;
					if(res[num3-num]==0)
						res.erase(res.find(num3-num));
					res[num3-num2]++;
					gg += num3-num2;
				}
			}
			temp.erase(temp.lower_bound(x));
		}
		else{
			// add a element
			if(temp.size()==0){
				cout<<gg<<endl;
				temp.insert(x);
				continue;
			}
			int num = *pos;
			if(pos==temp.begin()){
				res[*pos-x]++;
				gg += *pos-x;
			}else if(pos==temp.end()){
				res[x-*--pos]++;
				gg += x - *pos;
			}else{
				int num2 = *--pos;
				res[x-num2]++;
				res[num-x]++;
				res[num-num2]--;
				if(res[num-num2]==0)
					res.erase(res.find(num-num2));
			}
			temp.insert(x);
		}
		ans = gg;
		if(res.size())
			ans -= (*(--res.end())).fi;
		cout<<ans<<endl;
		
	}
	return 0;
}