Codeforces Round 916 (Div. 3)(A~E2)

A

统计一下每个字母的出现次数然后输出即可

#include <bits/stdc++.h>
#define rep(i,a,b) for(register int i = (a); i <= (b); ++i)
#define fep(i,a,b) for(register int i = (a); i >= (b); --i)
#define ls p<<1
#define rs p<<1|1
#define PII pair<int, int>
#define ll long long
#define ull unsigned long long
#define db double
#define endl '\n'
#define debug(a) cout<<#a<<"="<<a<<endl;
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define INF 0x3f3f3f3f 
using namespace std;
const int N = 1e5+10;
int t; 
void solve()
{
	int n; string str;
	cin >> n >> str;
	map<char,int>cnt;
	rep(i,0,str.size()-1)
	{
		cnt[str[i]]++;
	}
	int ans = 0;
	for(auto x:cnt)
	{
		int num=x.first-'A'+1;
		if(x.second>=num)	ans++;
	}
	cout << ans << endl;
}
int main()
{
	IOS
//  	freopen("1.in", "r", stdin);
	cin >> t;
	while(t --)
	solve();
	return 0;
}

B

前k个数是后n-k个数的顺序，后面的数逆序即可

#include <bits/stdc++.h>
#define rep(i,a,b) for(register int i = (a); i <= (b); ++i)
#define fep(i,a,b) for(register int i = (a); i >= (b); --i)
#define ls p<<1
#define rs p<<1|1
#define PII pair<int, int>
#define ll long long
#define ull unsigned long long
#define db double
#define endl '\n'
#define debug(a) cout<<#a<<"="<<a<<endl;
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define INF 0x3f3f3f3f 
using namespace std;
const int N = 1e5+10;
int t,a[60]; 
void solve()
{
	int n, k, cnt=1; cin >> n >> k;
	rep(i,1,n)	a[i]=0;
	rep(i,n-k,n)	a[cnt++] = i;
	fep(i,n-k-1,1)	a[cnt++] = i;
	rep(i,1,n)	cout << a[i] << ' ';
	cout << endl;
}
int main()
{
	IOS
//  	freopen("1.in", "r", stdin);
	cin >> t;
	while(t --)
	solve();
	return 0;
}

C

#include <bits/stdc++.h>
#define rep(i,a,b) for(register int i = (a); i <= (b); ++i)
#define fep(i,a,b) for(register int i = (a); i >= (b); --i)
#define ls p<<1
#define rs p<<1|1
#define PII pair<int, int>
#define ll long long
#define ull unsigned long long
#define db double
#define endl '\n'
#define debug(a) cout<<#a<<"="<<a<<endl;
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define INF 0x3f3f3f3f 
using namespace std;
const int N = 2e5+10;
int t;
int a[N], b[N], s[N];
void solve()
{
	int n, k;	cin >> n >> k;
	rep(i,1,n)	s[i]=0;
	rep(i,1,n)	cin >> a[i], s[i] = s[i-1]+a[i];
	rep(i,1,n)	cin >> b[i];
	int maxx=-INF, ans=-INF;
	rep(i,1,min(k,n))
	{
		maxx=max(maxx, b[i]);
		int cur=s[i]; cur+=max(0,k-i)*maxx;
		ans=max(cur,ans);
	}
	cout << ans << endl;
}
int main()
{
	IOS
//  	freopen("1.in", "r", stdin);
	cin >> t;
	while(t --)
	solve();
	return 0;
}

D

考虑维护每一个数组前缀和后缀的最大值

然后考虑3种情况即那个数在中间

a、b、c均可以在中间，

a在中间有b a c和c a b两种情况

b在中间有a b c和c b a两种情况

c在中间有a c b和b a c两种情况

这里只说a在中间的情况

我们枚举a的位置\(i\in[2,n-1]\)

那么对于b a c，b在a左边所以就是b应选择的是\([1,i-1]\)中的最大值，也就是lb[i-1]

c在右边应选择的就是rc[i+1]

这种情况的最大值就是\(a[i]+lb[i-1]+rc[i+1]\)

然后对其他几种情况做同样的处理即可

#include <bits/stdc++.h>
#define rep(i,a,b) for(register int i = (a); i <= (b); ++i)
#define fep(i,a,b) for(register int i = (a); i >= (b); --i)
#define ls p<<1
#define rs p<<1|1
#define PII pair<int, int>
#define ll long long
#define ull unsigned long long
#define db double
#define endl '\n'
#define debug(a) cout<<#a<<"="<<a<<endl;
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define INF 0x3f3f3f3f 
using namespace std;
const int N = 2e5+10;
int t, a[N], b[N], c[N];
int la[N], ra[N], lb[N], rb[N], lc[N], rc[N];
void solve()
{
	int n; cin >> n;
	rep(i,0,n+1)	la[i]=0,ra[i]=0,lb[i]=0,rb[i]=0,lc[i]=0,rc[i]=0;
	rep(i,1,n)	cin >> a[i];
	rep(i,1,n)	cin >> b[i];
	rep(i,1,n)	cin >> c[i];
	rep(i,1,n)	la[i] = max(la[i-1], a[i]);
	fep(i,n,1)	ra[i] = max(ra[i+1], a[i]);
	rep(i,1,n)	lb[i] = max(lb[i-1], b[i]);
	fep(i,n,1)	rb[i] = max(rb[i+1], b[i]);
	rep(i,1,n)	lc[i] = max(lc[i-1], c[i]);
	fep(i,n,1)	rc[i] = max(rc[i+1], c[i]);
	ll ans=0;
	rep(i,2,n-1)
	{
		ll maxx=0;
		maxx = max(maxx, (ll)a[i] + lb[i-1] + rc[i+1]);
		maxx = max(maxx, (ll)a[i] + lc[i-1] + rb[i+1]);
		maxx = max(maxx, (ll)b[i] + la[i-1] + rc[i+1]);
		maxx = max(maxx, (ll)b[i] + lc[i-1] + ra[i+1]);
		maxx = max(maxx, (ll)c[i] + la[i-1] + rb[i+1]);
		maxx = max(maxx, (ll)c[i] + lb[i-1] + ra[i+1]);
		ans = max(ans, maxx);
	}
	cout << ans << endl;
}
int main()
{
	IOS
//  	freopen("1.in", "r", stdin);
	cin >> t;
	while(t --)
	solve();
	return 0;
}

E1、E2

这道题目从数学的角度推公式比较好理解

考虑A、B选择的策略

考虑两列

a:x,y

b:u,v

a选择前面的得到的结果是x-1-(v-1)=x-v

a选择后面的得到的结果是y-1-(u-1)=y-u

哪个更优选择哪个

\((x-v)>(y-u) \rightarrow x+u > y+v也就是每个人会选择当前a[i]+b[i]\)值较大的这一列进行操作

#include <bits/stdc++.h>
#define rep(i,a,b) for(register int i = (a); i <= (b); ++i)
#define fep(i,a,b) for(register int i = (a); i >= (b); --i)
#define ls p<<1
#define rs p<<1|1
#define PII pair<int, int>
#define ll long long
#define ull unsigned long long
#define db double
#define endl '\n'
#define debug(a) cout<<#a<<"="<<a<<endl;
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define INF 0x3f3f3f3f 
using namespace std;
const int N = 2e5+10;
int t;
struct wy
{
	ll x, y, sum;
}a[N];
bool cmp(wy a, wy b)
{
	return a.sum > b.sum;
}
void solve()
{
	int n;	cin >> n;
	rep(i,1,n)	cin >> a[i].x;
	rep(i,1,n)	cin >> a[i].y, a[i].sum=a[i].x+a[i].y;
	sort(a+1,a+1+n, cmp);
	ll ans=0;
	rep(i,1,n)
	{
		if(i&1)	ans += a[i].x-1;
		else	ans -= a[i].y-1;
	}
	cout << ans << endl;
}
int main()
{
	IOS
//  	freopen("1.in", "r", stdin);
	cin >> t;
	while(t --)
	solve();
	return 0;
}