Codeforces Round #735 (Div. 2) 题解

A

发现长度为 (2) 的一定是最好的。

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;
#define endl '
'
#define debug(x) cerr << #x << ": " << x << endl
#define pb(a) push_back(a)
#define set0(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dwn(i,a,b) for(int i=(a);i>=(b);i--)
#define ceil(a,b) (a+(b-1))/b
#define INF 0x3f3f3f3f
#define ll_INF 0x7f7f7f7f7f7f7f7f
typedef long long ll;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;

#define int long long

inline void read(int &x) {
    int s=0;x=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-')x=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') s=(s<<3)+(s<<1)+ch-'0',ch=getchar();
    x*=s;
}

signed main(){
	int T; cin>>T;
	while(T--){
		int n; cin>>n;
		
		int a=0, b=0, res=0;
		rep(i,1,n){
			a=b; cin>>b;
			res=max(res, a*b);
		}
		cout<<res<<endl;
	}	
    return 0;
}

B

暴力

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;
#define endl '
'
#define debug(x) cerr << #x << ": " << x << endl
#define pb(a) push_back(a)
#define set0(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dwn(i,a,b) for(int i=(a);i>=(b);i--)
#define ceil(a,b) (a+(b-1))/b
#define INF 0x3f3f3f3f
#define ll_INF 0x7f7f7f7f7f7f7f7f
typedef long long ll;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;

inline void read(int &x) {
    int s=0;x=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-')x=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') s=(s<<3)+(s<<1)+ch-'0',ch=getchar();
    x*=s;
}

const int N=3e5+5;

int w[N];

int main(){
	int T; cin>>T;
	while(T--){
		int n, k; read(n), read(k);
		rep(i,1,n) read(w[i]);
		
		ll res=-INF;
		rep(i, max(1, n-2000), n) rep(j, i+1, n){
			res=max(res, 1LL*i*j-1LL*k*(w[i]|w[j]));
		}
		cout<<res<<endl;
	}	
    return 0;
}

C

什么样的数 (k) 一定会出现在序列 (noplus 0 ... noplus m) 中？

换句话说，就是 (k = noplus x)，其中 (xin [0,m])。

进一步得到应有 (noplus k leq m)，这样的 (k) 一定会出现在序列中。

我们的目标是找到一个最小的 (k) 使得 (noplus k geq m+1)。

从而转化为一个经典问题：给出 (a, c) 求最小的 (b) 使得： (a oplus b geq c)

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;

#define endl '
'
#define debug(x) cerr << #x << ": " << x << endl
#define pb(a) push_back(a)
#define set0(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dwn(i,a,b) for(int i=(a);i>=(b);i--)
#define ceil(a,b) (a+(b-1))/b

#define all(x) (x).begin(), (x).end()
#define SUM(a) accumulate(all(a), 0LL)
#define MIN(a) (*min_element(all(a)))
#define MAX(a) (*max_element(all(a)))
#define lb(a, x) distance(begin(a), lower_bound(all(a), (x)))
#define ub(a, x) distance(begin(a), upper_bound(all(a), (x)))

#define INF 0x3f3f3f3f
#define ll_INF 0x7f7f7f7f7f7f7f7f

using pii = pair<int, int>;
using pdd = pair<double, double>;
using vi = vector<int>;
using vvi = vector<vi>;
using vb = vector<bool>;
using vpii = vector<pii>;
using ll = long long;
using ull = unsigned long long;

inline void read(int &x) {
    int s=0;x=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-')x=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') s=(s<<3)+(s<<1)+ch-'0',ch=getchar();
    x*=s;
}

int main(){
	int T; cin>>T;
	while(T--){
		int n, m; read(n), read(m);
		int a=n, b=0, c=m+1;
		
		dwn(i,30,0){
			int x=a>>i&1, y=c>>i&1;	
			if(y && !x) b+=(1<<i);

			if((a^b)>=c) break;
		}
		cout<<b<<endl;
	}
    return 0;
}

D

构造，就在中间插个 b，然后如果是奇数就在尾部插个 c，其它全部选 a，见代码：

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;

#define endl '
'
#define debug(x) cerr << #x << ": " << x << endl
#define pb(a) push_back(a)
#define set0(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dwn(i,a,b) for(int i=(a);i>=(b);i--)
#define ceil(a,b) (a+(b-1))/b

#define all(x) (x).begin(), (x).end()
#define SUM(a) accumulate(all(a), 0LL)
#define MIN(a) (*min_element(all(a)))
#define MAX(a) (*max_element(all(a)))
#define lb(a, x) distance(begin(a), lower_bound(all(a), (x)))
#define ub(a, x) distance(begin(a), upper_bound(all(a), (x)))

#define INF 0x3f3f3f3f
#define ll_INF 0x7f7f7f7f7f7f7f7f

using pii = pair<int, int>;
using pdd = pair<double, double>;
using vi = vector<int>;
using vvi = vector<vi>;
using vb = vector<bool>;
using vpii = vector<pii>;
using ll = long long;
using ull = unsigned long long;

inline void read(int &x) {
    int s=0;x=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-')x=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') s=(s<<3)+(s<<1)+ch-'0',ch=getchar();
    x*=s;
}

const int N=1e5+5;
char res[N];

int main(){
	int T; cin>>T;
	while(T--){
		int n; read(n);
		
		rep(i,1,n){
			res[i]='a';
			if(i==(n+1)/2) res[i]='b';
		}
		if(n&1) res[n]='c';
		
		rep(i,1,n) cout<<res[i];
		cout<<endl;
	}
    return 0;
}