Educational Codeforces Round 80 (Rated for Div. 2)题解

A. Deadline
$x+\lceil \frac{d}{x+1}\rceil$ 我们就可以想到基本不等式，即取x/2时不等式取到最大值，于是就可以得出答案。

#include
using namespace std;
typedef long long int ll;
int main(){
	int T;scanf("%d",&T);
	while(T--){
		ll d,n;scanf("%lld %lld",&n,&d);
		if(n & 1) n = (n / 2 + 1) * (n / 2 + 1);
		else n = (n / 2) * (n / 2 + 1);
		if(n >= d) puts("YES");
		else puts("NO");
	}
}

B. Yet Another Meme Problem
遇事不决先打表，很容易发现只有m为9，99，999，9999，并且n<=m时成立，于是就。。。。

#include
using namespace std;
int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		ll n,m;scanf("%lld %lld",&n,&m);
		ll ans = 0,res = 1;
		for(int i = 1;;i++){
			res *= 10;
			if(res - 1 <= m) ans++;
			else break;
		}
		printf("%lld\n",ans * n);
	}
}

C. Two Arrays
给你n个数，m个位置，排成两个序列，一个不递减，一个不递增，问你一共有几中组合方式。
那我用dp[i][j] 表示i位长的串以j结尾的个数，起点为dp[0][n - 1],那前一位对后一位的贡献就是dp[i + 1][k] += dp[i][j] * (j - k + 1),结束得值为dp[m][0~n];

#include
typedef unsigned long long int ull;
typedef long long int ll;
const int    maxn = 1e6 + 5;
const int    maxm = 1e5 + 10;
const int    mod  = 1e9 + 7;
const ll     INF  = 1e18 + 100;
const int    inf  = 0x3f3f3f3f;
const double pi   = acos(-1.0);
const double eps  = 1e-6;
inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}return x*f;}
inline ll  readll(){ll x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}return x*f;}
ll eular(ll n){ll ans = n;for(int i = 2; i*i <= n; i++){if(n % i == 0){ans -= ans/i;while(n % i == 0) n /= i;}}if(n > 1)ans -= ans/n; return ans;}
ll qpow(ll a, ll n){ll ans = 1;while(n){if(n&1){ans = (ans * a) % mod;}a = (a * a) % mod;n >>= 1;}return ans;}
ll gcd(ll a,ll b){return b>0?gcd(b,a%b):a;}
using namespace std;
ll dp[11][1010];
int main(){
	int n,m;scanf("%d %d",&n,&m);
	dp[0][n - 1] = 1;
	for(int i = 0;i < m;i++){
		for(int j = 0;j <= n;j++){
			ll res = 1;
			for(int k = j;k >= 0;k--){
				dp[i + 1][k] += dp[i][j] * res % mod;
				dp[i + 1][k] %= mod;
				res++;
			}
		}
	}
	ll ans = 0;
	for(int i = 0;i <= n;i++) ans += dp[m][i],ans %= mod;
	printf("%lld\n",ans);
}

D. Minimax Problem
给你n行，每行m个数，让你选出任意两行，使得两行组合后得最小值最大。组合得定义给同一列取较大得那个数。
因为m只有8，考虑二分答案，check中，将每行所有得情况全部暴力计算出来，然后对于这些情况$((1<<m)-1{)}^2$得复杂度暴力计算出来就好了，判断条件为$((i|j)==((1<<m)-1))$。

#include
#define accept return 0
#define  mes(a, b)  memset(a, b, sizeof a)
typedef unsigned long long int ull;
typedef long long int ll;
const int    maxn = 3e5 + 5;
const int    maxm = 1e5 + 10;
const int    mod  = 1e9 + 7;
const ll     INF  = 1e18 + 100;
const int    inf  = 0x3f3f3f3f;
const double pi   = acos(-1.0);
const double eps  = 1e-6;
inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}return x*f;}
inline ll  readll(){ll x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}return x*f;}
ll eular(ll n){ll ans = n;for(int i = 2; i*i <= n; i++){if(n % i == 0){ans -= ans/i;while(n % i == 0) n /= i;}}if(n > 1)ans -= ans/n; return ans;}
ll qpow(ll a, ll n){ll ans = 1;while(n){if(n&1){ans = (ans * a) % mod;}a = (a * a) % mod;n >>= 1;}return ans;}
ll gcd(ll a,ll b){return b>0?gcd(b,a%b):a;}
using namespace std;
int ans1,ans2;
int used[500];
int tot;
int n,m;
int mp[maxn][20];
bool check(int pos){
	for(int i = 0;i <= tot;i++) used[i] = 0;
	for(int i = 1;i <= n;i++){
		int tmp = 0;
		for(int j = 1;j <= m;j++){
			if(mp[i][j] >= pos) tmp = tmp << 1 | 1;
			else tmp = tmp << 1;
		}
		used[tmp] = i;
	} 
	for(int i = 0;i <= tot;i++) for(int j = 0;j <= tot;j++){
		if((i | j) == (tot) && used[i] && used[j]){
			ans1 = used[i];ans2 = used[j];
			return true;
		}
	}
	return false;
}
int main(){
	scanf("%d %d",&n,&m);
	tot = (1 << m) - 1;	 
	for(int i = 1;i <= n;i++) for(int j = 1;j <= m;j++){
		scanf("%d",&mp[i][j]);
	}
	int l = 0,r = 1e9,ans = 0;
	while(l <= r){
		int mid = l + r >> 1;
		if(check(mid)) ans = mid,l = mid + 1;
		else r = mid - 1;
	}
	printf("%d %d\n",ans1,ans2);
}