CF1288

A

考虑(x + 1 = sqrt{d})时在有理域上有最优界。
那我在整数域上附近取三个点取min就行了。

// code by fhq_treap
#include<bits/stdc++.h>
#define ll long long
#define N 300005

inline ll read(){
    char C=getchar();
    ll A=0 , F=1;
    while(('0' > C || C > '9') && (C != '-')) C=getchar();
    if(C == '-') F=-1 , C=getchar();
    while('0' <= C && C <= '9') A=(A << 1)+(A << 3)+(C - 48) , C=getchar();
    return A*F;
}

template <typename T>
void write(T x)
{
    if(x < 0) {
        putchar('-');
        x = -x;
    }
    if(x > 9)
        write(x/10);
    putchar(x % 10 + '0');
    return;
}

int n,d,t;

int main(){
	scanf("%d",&t);
	while(t -- ){
		scanf("%d%d",&n,&d);
		int ans = d;
		int k = sqrt(d);
		ans = std::min(ans,(d + k - 1) / k + k - 1);
		k ++ ;
		ans = std::min(ans,(d + k - 1) / k + k - 1);
		k -= 2;
		if(k != 0)
		ans = std::min(ans,(d + k - 1)/ k + k - 1);
		puts((ans <= n) ? "Yes" : "No");
	}
}

B

考虑枚举(b)的位数，然后发现(b)一定是 (9999) 这类的形式，而(a)可以随意取。

// code by fhq_treap
#include<bits/stdc++.h>
#define ll long long
#define N 300005

inline ll read(){
    char C=getchar();
    ll A=0 , F=1;
    while(('0' > C || C > '9') && (C != '-')) C=getchar();
    if(C == '-') F=-1 , C=getchar();
    while('0' <= C && C <= '9') A=(A << 1)+(A << 3)+(C - 48) , C=getchar();
    return A*F;
}

template <typename T>
void write(T x)
{
    if(x < 0) {
        putchar('-');
        x = -x;
    }
    if(x > 9)
        write(x/10);
    putchar(x % 10 + '0');
    return;
}

int t;

int main(){
	scanf("%d",&t);
	while(t -- ){
		int a,b;
		scanf("%d%d",&a,&b);
		int k = 0,now = 9;
		while(b >= now){
            k++;
            now = now * 10 + 9;
		}
		std::cout<<1ll * k * a<<std::endl;
	}
}

C

直接暴力(dp),再利用前缀和优化到(O(n^2))

// code by fhq_treap
#include<bits/stdc++.h>
#define ll long long
#define N 300005
#define mod ((ll)1e9 + 7)

inline ll read(){
    char C=getchar();
    ll A=0 , F=1;
    while(('0' > C || C > '9') && (C != '-')) C=getchar();
    if(C == '-') F=-1 , C=getchar();
    while('0' <= C && C <= '9') A=(A << 1)+(A << 3)+(C - 48) , C=getchar();
    return A*F;
}

template <typename T>
void write(T x)
{
    if(x < 0) {
        putchar('-');
        x = -x;
    }
    if(x > 9)
        write(x/10);
    putchar(x % 10 + '0');
    return;
}
//f[i][j][k] = sum(f[i - 1][p][q](q <= k,p >= j,j <= k))
int f[1005][1005];
int g[1005][1005];
int n,m;
inline void got(){
	int tmp = 0;
	for(int i = 1;i <= n;++i)
	for(int j = 1;j <= n;++j)
	g[i][j] = 0;
	for(int i = 1;i <= n;++i){
		tmp = 0;
		for(int j = n;j >= 1;--j){
			tmp = (tmp + f[i][j]) % mod;
			g[i][j] = g[i - 1][j];
			g[i][j] = (g[i][j] + tmp) % mod;
		}
	}
}
int main(){
	scanf("%d%d",&n,&m);
	f[1][n] = 1;
	got();
	for(int i = 1;i <= m;++i){
		for(int j = 1;j <= n;++j){
			for(int k = j;k <= n;++k){
				f[j][k] = g[j][k];
			}
		}
		got();
	}
	std::cout<<g[n][1] % mod<<std::endl;
}

D

二分答案，将一个长度为 (m) 的数字序列转换为 (01) 序列。

可以获得这个答案则有两个(01)串或起来为全1串。

可以(O({2^m}^2))做。

// code by fhq_treap
#include<bits/stdc++.h>
#define ll long long
#define N 300005

inline ll read(){
    char C=getchar();
    ll A=0 , F=1;
    while(('0' > C || C > '9') && (C != '-')) C=getchar();
    if(C == '-') F=-1 , C=getchar();
    while('0' <= C && C <= '9') A=(A << 1)+(A << 3)+(C - 48) , C=getchar();
    return A*F;
}

template <typename T>
void write(T x)
{
    if(x < 0) {
        putchar('-');
        x = -x;
    }
    if(x > 9)
        write(x/10);
    putchar(x % 10 + '0');
    return;
}

int n,m;
int a[300005][8];

int ansx = 1,ansy  = 1;
int t[(1 << 9)];

inline bool check(int lim){
	for(int i = 0;i < (1ll << m);++i)
	t[i] = 0;
	for(int i = 1;i <= n;++i){
		int tmp = 0;
		for(int j = 1;j <= m;++j){
			if(a[i][j] >= lim)
			tmp = tmp | (1ll << (j - 1));
		}
		t[tmp] = i;
	}
	for(int i = 0;i < (1ll << m);++i)
	for(int j = 0;j < (1ll << m);++j){
		if(((i | j) == ((1ll << m) - 1)) && (t[i]) && t[j]){
			ansx = t[i];
			ansy = t[j];
			return 1;
		}
	}
	return 0;
}

int main(){
	scanf("%d%d",&n,&m);
	for(int i = 1;i <= n;++i)
	for(int j = 1;j <= m;++j){
		scanf("%d",&a[i][j]);
	}
	int l = -1,r = 1e9;
	#define mid ((l + r) >> 1)
	while(l <= r){
		if(check(mid))
		l = mid + 1;
		else
		r = mid - 1;
	}
	l --;
	while(check(l + 1))
	l ++ ;
	std::cout<<ansx<<" "<<ansy<<std::endl;
}

E

经典套路，前面留(m)个位置，用树状数组维护排名。

// code by fhq_treap
#include<bits/stdc++.h>
#define ll long long
#define N 300005

inline ll read(){
    char C=getchar();
    ll A=0 , F=1;
    while(('0' > C || C > '9') && (C != '-')) C=getchar();
    if(C == '-') F=-1 , C=getchar();
    while('0' <= C && C <= '9') A=(A << 1)+(A << 3)+(C - 48) , C=getchar();
    return A*F;
}

template <typename T>
void write(T x)
{
    if(x < 0) {
        putchar('-');
        x = -x;
    }
    if(x > 9)
        write(x/10);
    putchar(x % 10 + '0');
    return;
}

int n;
int a[N];
int minn[N],maxx[N];
int T[N << 2];
#define lowbit(x) (x & -x)

inline void add(int x,int p){
	for(int i = x;i <= N << 2;i += lowbit(i))
	T[i] += p;
}

inline int find(int x){
	int ans = 0;
	for(int i = x;i;i -= lowbit(i))
	ans += T[i];
	return ans;
}
int m,pos[N];
int main(){
	scanf("%d%d",&n,&m);
	for(int i = 1;i <= n;++i)
	minn[i] = maxx[i] = i,pos[i] = m + i,add(pos[i],1);
	int now = m;
	for(int i = 1;i <= m;++i){
		int k;
		scanf("%d",&k);
		minn[k] = 1;
		maxx[k] = std::max(maxx[k],find(pos[k]));
		add(pos[k],-1);
		pos[k] = now -- ;
		add(pos[k],1);
	}
	for(int i = 1;i <= n;++i){
		maxx[i] = std::max(maxx[i],find(pos[i]));
		std::cout<<minn[i]<<" "<<maxx[i]<<std::endl;
	}
}