Codeforces Round 927 (Div. 3)（A~F）

目录

• A
• B
• C
• D
• E
• F

A

第一个遇到连续两个荆棘的地方就不能再赢金币了。

所以统计连续两个荆棘之前的所有金币

#include <bits/stdc++.h>
#define int long long
#define rep(i,a,b) for(int i = (a); i <= (b); ++i)
#define fep(i,a,b) for(int i = (a); i >= (b); --i)
#define pii pair<int, int>
#define ll long long
#define db double
#define endl '\n'
#define x first
#define y second
#define pb push_back

using namespace std;

void solve() {
	int n;
	cin>>n;
	string str;
	cin>>str;
	int pos=n;
	rep(i,0,n-2){
		if(str[i]=='*'&&str[i+1]=='*'){
			pos=i;
			break;
		}
	}
	int ans=0;
	if(pos!=n){
		rep(i,0,pos-1){
			if(str[i]=='@'){
				ans++;
			}
		}
	}else{
		rep(i,0,n-1){
			if(str[i]=='@'){
				ans++;
			}
		}
	}
	cout<<ans<<endl;
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
//	freopen("1.in", "r", stdin);
	int _;
	cin>>_;
	while(_--)
	solve();
	return 0;
}

B

最后需要满足\(a[i]>a[i-1]\)

\(a[i]\)只能增加自身的倍数

只需要计算\(a[i]\)最终会变为自己的多少倍会严格大于\(a[i-1]\),当\(a[i-1]\)是\(a[i]\)的倍数的时候，必须再\(+1\)才能保证严格大于。

#include <bits/stdc++.h>
#define int long long
#define rep(i,a,b) for(int i = (a); i <= (b); ++i)
#define fep(i,a,b) for(int i = (a); i >= (b); --i)
#define pii pair<int, int>
#define ll long long
#define db double
#define endl '\n'
#define x first
#define y second
#define pb push_back

using namespace std;

void solve() {
	int n;
	cin>>n;
	vector<int>a(n+1);
	rep(i,1,n){
		cin>>a[i];
	}
	rep(i,1,n){
		if(a[i]<=a[i-1]){
			int cnt=a[i-1]/a[i]+1;
			a[i]*=cnt;
		}
	}

	cout<<a[n]<<endl;
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
//	freopen("1.in", "r", stdin);
	int _;
	cin>>_;
	while(_--)
	solve();
	return 0;
}

C

正着算会溢出

考虑倒着算，也就是先算最后一个留下的，然后边乘边模

#include <bits/stdc++.h>
#define int long long
#define rep(i,a,b) for(int i = (a); i <= (b); ++i)
#define fep(i,a,b) for(int i = (a); i >= (b); --i)
#define pii pair<int, int>
#define ll long long
#define db double
#define endl '\n'
#define x first
#define y second
#define pb push_back

using namespace std;

void solve() {
	int n,m;
	cin>>n>>m;
	struct node{
		int val,pos;
		bool operator<(const node&t)const{
			return pos>t.pos;
		}
	};
	vector<node>a(n+1);
	rep(i,1,n){
		cin>>a[i].val;
	}
	string str;
	cin>>str;

	int cnt=1;
	int l=1,r=n;

	rep(i,0,str.size()-1){
		if(str[i]=='L'){
			a[l].pos=cnt++;
			l++;
		}else{
			a[r].pos=cnt++;
			r--;
		}
	}

	sort(a.begin()+1,a.end());
//	rep(i,1,n){
//		cout<<a[i].val<<' '<<a[i].pos<<endl;
//	}
	vector<int>ans;
	int mul=1;
	rep(i,1,n){
		int k=a[i].val*mul%m;
		ans.pb(k);
		mul=k;
	}
	fep(i,ans.size()-1,0){
		cout<<ans[i]<<' ';
	}
	cout<<endl;
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
//	freopen("1.in", "r", stdin);
	int _;
	cin>>_;
	while(_--)
	solve();
	return 0;
}

D

模拟题.

同花色的两两配对，是奇数的话，再用一张王去配对

判断有没有解是通过奇数牌的张数和王的张数判断。

#include <bits/stdc++.h>
#define int long long
#define rep(i,a,b) for(int i = (a); i <= (b); ++i)
#define fep(i,a,b) for(int i = (a); i >= (b); --i)
#define pii pair<int, int>
#define ll long long
#define db double
#define endl '\n'
#define x first
#define y second
#define pb push_back

using namespace std;

void solve() {
	int n;
	cin>>n;
	//0:C,1:D,2:H,3:S
	set<int>s[4];
	char c;
	cin>>c;
	map<char,int>mp;
	mp['C']=0;
	mp['D']=1;
	mp['H']=2;
	mp['S']=3;

	map<int,char>pm;
	pm[0]='C';
	pm[1]='D';
	pm[2]='H';
	pm[3]='S';
	rep(i,1,2*n){
		string str;
		cin>>str;
		char k=str[1];
		int num=str[0]-'0';
		if(k=='C'){
			s[0].insert(num);
		}else if(k=='D'){
			s[1].insert(num);
		}else if(k=='H'){
			s[2].insert(num);
		}else{
			s[3].insert(num);
		}
	}
	int kk=0;
	rep(i,0,3){
		if(mp[c]==i){
			continue;
		}
		kk+=s[i].size()%2;
	}
	if(kk>s[mp[c]].size()){
		cout<<"IMPOSSIBLE"<<endl;
		return;
	}

	rep(i,0,3){
		if(mp[c]==i){
			continue;
		}
		int ji=s[i].size()%2;
		if(ji==1){
			cout<<*s[i].begin()<<pm[i];
			s[i].erase(s[i].begin());
			cout<<' ';
			cout<<*s[mp[c]].begin()<<pm[mp[c]];
			s[mp[c]].erase(s[mp[c]].begin());
			cout<<endl;
			while(s[i].size()>=2){
				cout<<*s[i].begin()<<pm[i];
				s[i].erase(s[i].begin());
				cout<<' ';
				cout<<*s[i].begin()<<pm[i];
				s[i].erase(s[i].begin());
				cout<<endl;
			}
		}else{
			while(s[i].size()>=2){
				cout<<*s[i].begin()<<pm[i];
				s[i].erase(s[i].begin());
				cout<<' ';
				cout<<*s[i].begin()<<pm[i];
				s[i].erase(s[i].begin());
				cout<<endl;
			}
		}
	}
	while(s[mp[c]].size()>=2){
		cout<<*s[mp[c]].begin()<<pm[mp[c]]<<' ';
		s[mp[c]].erase(s[mp[c]].begin());
		cout<<*s[mp[c]].begin()<<pm[mp[c]]<<' ';
		s[mp[c]].erase(s[mp[c]].begin());
		cout<<endl;
	}
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
//	freopen("1.in", "r", stdin);
	int _;
	cin>>_;
	while(_--)
	solve();
	return 0;
}

E

数学题。

考虑每一位对答案的贡献

\(12345\)

个位会对答案贡献\(12345\)

十位会对答案贡献\(1234\)

百位会对答案贡献\(123\)

千位会对答案贡献\(12\)

万位会对答案贡献\(1\)

然后会发现规律，将上面列竖式相加

每一位的结果就是当前位的数和前面所有数的和，然后倒着处理一下进位。

#include <bits/stdc++.h>
#define int long long
#define rep(i,a,b) for(int i = (a); i <= (b); ++i)
#define fep(i,a,b) for(int i = (a); i >= (b); --i)
#define pii pair<int, int>
#define ll long long
#define db double
#define endl '\n'
#define x first
#define y second
#define pb push_back

using namespace std;

void solve() {
	int n;
	cin>>n;
	string str;
	cin>>str;
	vector<int>s(n+1,0);
	rep(i,1,n){
		s[i]=s[i-1]+str[i-1]-'0';
	}
	fep(i,n,1){
		s[i-1]+=s[i]/10;
		s[i]%=10;
	}
	bool flag=false;
	rep(i,0,n){
		if(s[i]||flag){
			cout<<s[i];
			flag=true;
		}
	}
	cout<<endl;
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
//	freopen("1.in", "r", stdin);
	cout.tie(0);
	int _;
	cin>>_;
	while(_--)
	solve();
	return 0;
}

F

\(dp[i][0|1]\)：表示在第i个点不喂或喂猫，合法的能为猫的最大值

转移

\(nxt[i]\):在i处喂猫的话能喂猫的最有边的位置

\(cf[i]\):在i处喂猫，所能喂猫的所有数量

\(f[i][0]=max(f[i-1][0],f[i-1][1]);\)

\(f[i][1]=max(f[nxt[i]-1][1],f[nxt[i]-1][0])+cf[i];\)

\(cf[i]\)可以用差分处理一下

\(nxt[i]\)可以倒序枚举用\(nxt[i-1]\)去更新\(nxt[i]\)

#include <bits/stdc++.h>
#define int long long
#define rep(i,a,b) for(int i = (a); i <= (b); ++i)
#define fep(i,a,b) for(int i = (a); i >= (b); --i)
#define pii pair<int, int>
#define ll long long
#define db double
#define endl '\n'
#define x first
#define y second
#define pb push_back

using namespace std;

void solve() {
	int n,m;
	cin>>n>>m;
	vector<int>nxt(n+2),cf(n+2);
	rep(i,1,n){
		nxt[i]=i;
	}
	rep(i,1,m){
		int l,r;
		cin>>l>>r;
		cf[l]++;
		cf[r+1]--;
		nxt[r]=min(nxt[r],l);
	}

	//处理在每个点喂能为多少猫
	rep(i,1,n){
		cf[i]+=cf[i-1];
	}

	//处理转移的位置
	fep(i,n-1,1){
		nxt[i]=min(nxt[i],nxt[i+1]);
	}

	//dp
	vector<vector<int>>f(n+1,vector<int>(2));

	rep(i,1,n){
		f[i][0]=max(f[i-1][0],f[i-1][1]);
		f[i][1]=max(f[nxt[i]-1][1],f[nxt[i]-1][0])+cf[i];
	}
	cout<<max({f[n][0],f[n][1]})<<endl;
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
//	freopen("1.in", "r", stdin);
	int _;
	cin>>_;
	while(_--)
	solve();
	return 0;
}