Codeforces Round #735 (Div. 2) (A-D)

Posted 2022-11-28 TURNINING

传送门

A . 思路：对于一个很大的数，区间越大最小的数是越小的，他们的乘积也就越小，所以枚举相邻两个数就行了。

#include<bits/stdc++.h>
using namespace std;
 
#define lsn (u << 1)
#define rsn (u << 1 | 1)
#define mid (l + r >> 1)
 
typedef long long ll;
typedef unsigned long long ull;
 
typedef pair<double, double> P;
 
const int MAXN = 1e5 + 10;
const int MAX_LEN = 1e5 + 10;
const int MAX_LOG_V = 22;
const ll INF = 1e17;
const int inf = 0x3f3f3f3f;
const int mod = 19650827;
const double eps = 1e-7;
const ull B = 100000007;

int n;
ll a[MAXN];

void solve() 
	scanf("%d", &n);
	ll ans = 0;
	for(int i = 1; i <= n; i++) 
		scanf("%lld", &a[i]); 
	
	for(int i = 1; i < n; i++) 
		ans = max(min(a[i],a[i+1])*max(a[i],a[i+1]), ans);
	
	printf("%lld\\n", ans);



int main() 
	//ios::sync_with_stdio(false);
	int t = 1; scanf("%d", &t);  
	while(t--)  
		solve();
	

B： 思路：枚举后100个就行了。。。。

#include<bits/stdc++.h>
using namespace std;
 
#define lsn (u << 1)
#define rsn (u << 1 | 1)
#define mid (l + r >> 1)
 
typedef long long ll;
typedef unsigned long long ull;
 
typedef pair<double, double> P;
 
const int MAXN = 1e5 + 10;
const int MAX_LEN = 1e5 + 10;
const int MAX_LOG_V = 22;
const ll INF = 1e17;
const int inf = 0x3f3f3f3f;
const int mod = 19650827;
const double eps = 1e-7;
const ull B = 100000007; 
 
 
int n, k;
int a[MAXN];
 
void solve() 
	scanf("%d %d", &n, &k);
	for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
	ll ans = -1e17;
	for(int i = max(1, n-1000); i <= n; i++) 
		for(int j = i+1; j <= n; j++) 
			ans = max(ans, (ll)i*j-(ll)k*(a[i]|a[j]));
		
	
	printf("%lld\\n", ans);

 
 
int main() 
	//ios::sync_with_stdio(false);
	int t = 1; scanf("%d", &t);
	while(t--) 
		solve();
	
	return 0;

C 思路：观察发现$n>m$时 0 取不到，直接输出0。当$n<=m$时 令 $n⨁x=y$，根据题目要求，我们要找到一个y使$x>m$且 y 最小。$n⨁y>m$ , y $⨁$ n >= m + 1，怎么找最小的y呢，我们知道在二进制位中一个$2^i>{\sum }_{j=0}^{i-1}{2}^j$, 故只有当二进制位上n为0，m为1时我们才让y取1。

#include<bits/stdc++.h>
using namespace std;
 
#define lsn (u << 1)
#define rsn (u << 1 | 1)
#define mid (l + r >> 1)
 
typedef long long ll;
typedef unsigned long long ull;
 
typedef pair<double, double> P;
 
const int MAXN = 1e5 + 10;
const int MAX_LEN = 1e5 + 10;
const int MAX_LOG_V = 22;
const ll INF = 1e17;
const int inf = 0x3f3f3f3f;
const int mod = 19650827;
const double eps = 1e-7;
const ull B = 100000007; 


int n, m;


void solve() 
	scanf("%d %d", &n, &m);
	if(n > m) puts("0"); return ;
	m++;
	int res = 0;
	for(int i = 30; i >= 0; i--) 
		if(!(n >> i & 1) && (m >> i & 1)) res |= (1 << i);
		else if((n >> i & 1) && !(m >> i & 1)) break;
	
	printf("%d\\n", res);



int main() 
	//ios::sync_with_stdio(false);
	int t = 1; scanf("%d", &t);
	while(t--) 
		solve();
	
	return 0;

D 思路：研究一下样例发现，一连串含相同字符的字符串$(如aaaaa)$当再往后添加一个相同字符时，添加后相同长度的子串的个数为添加前的+1，一奇一偶相加肯定为奇数。特判一下n为1的情况

#include<bits/stdc++.h>
using namespace std;
 
#define lsn (u << 1)
#define rsn (u << 1 | 1)
#define mid (l + r >> 1)
 
typedef long long ll;
typedef unsigned long long ull;
 
typedef pair<double, double> P;
 
const int MAXN = 1e5 + 10;
const int MAX_LEN = 1e5 + 10;
const int MAX_LOG_V = 22;
const ll INF = 1e17;
const int inf = 0x3f3f3f3f;
const int mod = 19650827;
const double eps = 1e-7;
const ull B = 100000007; 


int n;


void solve() 
	scanf("%d", &n);
	if(n == 1) puts("a"); return ;
	if(n % 2 == 0) 
		int t = n / 2;
		for(int i = 1; i < t; i++) printf("a");
		printf("b");
		for(int i = 1; i <= t; i++) printf("a");
	
	else 
		int t = n / 2;
		for(int i = 1; i < t; i++) printf("a");
		printf("bc");
		for(int i = 1; i <= t; i++) printf("a");
	
	puts("");



int main() 
	//ios::sync_with_stdio(false);
	int t = 1; scanf("%d", &t);
	while(t--) 
		solve();
	
	return 0;