Solution Set -「CF 1514」

「CF 1514A」Perfectly Imperfect Array

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就看序列中是否存在不为平方数的元素即可。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
template<typename T>void sf(T &x){x=0;T f=0;char c=getchar();for(;c<'0'||c>'9';c=getchar())if(c=='-')f=1;for(;c>='0'&&c<='9';c=getchar())x=(x<<3)+(x<<1)+(c^'0');if(f)x=-x;}
template<typename T>void pf(T x,char l='\n'){static T s[100],t;if(x<0)putchar('-'),x=-x;do s[++t]=x%10,x/=10;while(x);while(t)putchar(s[t--]^'0');putchar(l);}
ll T,n,x;
bool check(ll x)
{
	for(ll i=1;i*i<=x;++i)	if(i*i==x)	return true;
	return false;
}
int main()
{
	for(sf(T);T;--T)
	{
		sf(n);
		bool fl=0;
		for(int i=1;i<=n;++i)
		{
			sf(x);
			if(!check(x))	fl=1;
		}
		if(fl)	puts("YES");
		else	puts("NO");
	}
	return 0;
}

「CF 1514B」AND 0, Sum Big

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和最大的情况就是每个数都只有一个 \(0\)（二进制下），于是答案就是 \(n^{k}\)。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
template<typename T>void sf(T &x){x=0;T f=0;char c=getchar();for(;c<'0'||c>'9';c=getchar())if(c=='-')f=1;for(;c>='0'&&c<='9';c=getchar())x=(x<<3)+(x<<1)+(c^'0');if(f)x=-x;}
template<typename T>void pf(T x,char l='\n'){static T s[100],t;if(x<0)putchar('-'),x=-x;do s[++t]=x%10,x/=10;while(x);while(t)putchar(s[t--]^'0');putchar(l);}
const int MOD=1e9+7;
int main()
{
	int T,n,k;
	for(sf(T);T;--T)
	{
		sf(n),sf(k);
		ll ans=1;
		for(int i=1;i<=k;++i)	ans*=n,ans%=MOD;
		pf(ans);
	}
	return 0;
}

「CF 1514C」Product 1 Modulo N

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把所有与 \(n\) 互质的数拉出来，如果 product 不满足要求，就把最后一个剔除。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
template<typename T>void sf(T &x){x=0;T f=0;char c=getchar();for(;c<'0'||c>'9';c=getchar())if(c=='-')f=1;for(;c>='0'&&c<='9';c=getchar())x=(x<<3)+(x<<1)+(c^'0');if(f)x=-x;}
template<typename T>void pf(T x,char l='\n'){static T s[100],t;if(x<0)putchar('-'),x=-x;do s[++t]=x%10,x/=10;while(x);while(t)putchar(s[t--]^'0');putchar(l);}
int n;
ll op(ll x,ll y){return x*y%n;}
int main()
{
	sf(n);
	std::vector<ll> ans;
	for(int i=1;i<=n;++i)	if(__gcd(i,n)==1)	ans.emplace_back(i);
	if(accumulate(ans.begin(),ans.end(),1,op)!=1)	ans.pop_back();
	pf(ans.size());
	for(int i:ans)	pf(i,' ');
	return 0;
}

「CF 1514D」Cut and Stick

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答案是 \(\max\{1,2x-l\}\)，\(l\) 为区间长度，\(x\) 为众数出现次数。

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>

using namespace std;

const int MAXN = 5e5 + 5, MAXM = 720 + 5;

char buf[1 << 21], *p1 = buf, *p2 = buf;
#define getchar( ) ( p1 == p2 && ( p2 = ( p1 = buf ) + fread( buf, 1, 1 << 21, stdin ), p1 == p2 ) ? EOF : *p1 ++ )

template<typename _T>
void read( _T &x ){
	x = 0; char c = getchar( ); _T f = 1;
	while( c < '0' || c > '9' ){ if( c == '-' )	f = -1; c = getchar( ); }
	while( c >= '0' && c <= '9' ){ x = ( x << 3 ) + ( x << 1 ) + ( c & 15 ); c = getchar( ); }
	x *= f;
}

template<typename _T>
void write( _T x ){
	if( x < 0 ){ putchar( '-' ); x = -x; }
	if( x > 9 ){ write( x / 10 ); }
	putchar( x % 10 + '0' );
}

template<typename _T>
void swapp( _T &one, _T &another ){ int temp = one; one = another; another = temp; }

template<typename _T>
_T MIN( _T one, _T another ){ return one > another ? another : one; }

template<typename _T>
_T MAX( _T one, _T another ){ return one > another ? one : another; }

int N, M;
int cube, each, kase, isa[MAXN], cnt[MAXN], pos[MAXN], vis[MAXN], bel[MAXN];
int lps[MAXM], rps[MAXM], App[MAXM][MAXM];
vector<int> disc, fur[MAXN];

int getID( int x ){ return lower_bound( disc.begin( ), disc.end( ), x ) - disc.begin( ) + 1; }

void build( ){
	memset( cnt, 0, sizeof( cnt ) );
	for( int i = 1; i <= cube; ++ i ){
		kase ++;
		for( int j = i; j <= cube; ++ j ){
			App[i][j] = App[i][j - 1];
			for( int k = lps[j]; k <= rps[j]; ++ k ){
				if( vis[isa[k]] != kase )	cnt[isa[k]] = 0;
				cnt[isa[k]] ++; App[i][j] = MAX( App[i][j], cnt[isa[k]] );
				vis[isa[k]] = kase;
			}
		}
	}
	memset( cnt, 0, sizeof( cnt ) );
}

int query( int opl, int opr ){
	if( bel[opl] == bel[opr] ){
		int res = 0; kase ++;
		for( int i = opl; i <= opr; ++ i ){
			if( vis[isa[i]] != kase )	cnt[isa[i]] = 0;
			cnt[isa[i]] ++; res = MAX( res, cnt[isa[i]] );
			vis[isa[i]] = kase;
		}
		return res;
	}
	int res = 0;
	res = App[bel[opl] + 1][bel[opr] - 1];
	for( int i = opl; i <= rps[bel[opl]]; ++ i ){
		int lim = fur[isa[i]].size( ) - 1;
		while( pos[i] + res <= lim && fur[isa[i]][pos[i] + res] <= opr )	res ++;
	}
	for( int i = lps[bel[opr]]; i <= opr; ++ i ){
		while( pos[i] - res >= 0 && fur[isa[i]][pos[i] - res] >= opl )	res ++;
	}
	return res;
}

signed main( ){
	read( N ); read( M ); each = 720; cube = ( N - 1 ) / each + 1;
	for( int i = 1; i <= N; ++ i ){ read( isa[i] ); disc.push_back( isa[i] ); }
	sort( disc.begin( ), disc.end( ) );
	disc.erase( unique( disc.begin( ), disc.end( ) ), disc.end( ) );
	for( int i = 1; i <= N; ++ i ){
		isa[i] = getID( isa[i] );
		fur[isa[i]].push_back( i );
		pos[i] = fur[isa[i]].size( ) - 1;
	}
	for( int i = 1; i <= cube; ++ i ){
		lps[i] = rps[i - 1] + 1; rps[i] = rps[i - 1] + each;
		if( i == cube )	rps[i] = N;
		for( int j = lps[i]; j <= rps[i]; ++ j )	bel[j] = i;
	}
	build( );
	int opl, opr;
	while( M -- > 0 ){
		read( opl ); read( opr );
		write( max( 1, 2 * query( opl, opr ) - ( opr - opl + 1 ) ) );
		putchar( '\n' );
	}
	return 0;
}

「CF 1514E」Baby Ehab's Hyper Apartment

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