Record - Nov. 20th, 2020 - Exam. SOL

LOC 2020.11.20 - Prob. 1

Desc. & Link.

\(C=2^{k}\bmod(a+b+c)\)

#include <cstdio>

typedef long long LL;

int A, B, C, K;

int Qkpow( int base, int indx, int mod ){
	int res = 1;
	while( indx ){
		if( indx & 1 )	res = ( LL )res * base % mod;
		base = ( LL )base * base % mod;
		indx >>= 1;
	}
	return res;
}

int main( ){
	int TC; scanf( "%d", &TC ); while( TC -- > 0 ){
		scanf( "%d%d%d%d", &A, &B, &C, &K );
		printf( "%lld\n", ( LL )C * Qkpow( 2, K, A + B + C ) % ( A + B + C ) );
	}
	return 0;
}

LOC 2020.11.20 - Prob. 2

Desc. & Link.

我先行否决 naive 的线段树。

输入暗示连边？

那就连吧。

考虑每一个连通块 \(S\)，如果连通块是个树，就只能选 \(|S|-1\) 个点。

如果存在环，那么每个数都取得到。

那么可以的情况就是询问区间包含的不是一棵树。

#include <cstdio>

const int MAXN = 2e5 + 5;

template<typename _T> _T MIN( const _T x, const _T y ){ return x > y ? y : x; }
template<typename _T> _T MAX( const _T x, const _T y ){ return x > y ? x : y; }

struct pointS{
	int one, ano;
	pointS( int O = 0, int A = 0 ){ one = A; ano = A; }
} pntS[MAXN];

struct starS{
	int to, nx;
	starS( int T = 0, int N = 0 ){ to = T; nx = N; }
} as[MAXN * 2];

int N, K, Q;
int totE;
int firS[MAXN], furS[MAXN], enS[MAXN], mxV[MAXN], mnV[MAXN], vis[MAXN];

void pushEdge( const int u, const int v ){ as[++ totE] = starS( v, firS[u] ); firS[u] = totE; furS[u] ++; }

void DFS( const int u, int &mxVt, int &mnVt, int &edgeS, int &nodeS ){
	mxVt = MAX( mxVt, u ); mnVt = MIN( mnVt, u );
	edgeS += furS[u]; nodeS ++; vis[u] = 1;
	for( int i = firS[u]; i; i = as[i].nx ){
		int v = as[i].to;
		if( vis[v] )	continue;
		DFS( v, mxVt, mnVt, edgeS, nodeS );
	}
}

int main( ) {
	scanf( "%d%d", &N, &K );
	for( int i = 1; i <= K; ++ i ){
		scanf( "%d%d", &pntS[i].one, &pntS[i].ano );
		pushEdge( pntS[i].one, pntS[i].ano );
		pushEdge( pntS[i].ano, pntS[i].one );
	}
	for( int i = 1; i <= N; ++ i )	enS[i] = N + 1;
	for( int i = 1; i <= N; ++ i ){
		if( vis[i] )	continue;
		int mxVt = 1, mnVt = N, edgeS = 0, nodeS = 0;
		DFS( i, mxVt, mnVt, edgeS, nodeS ); edgeS >>= 1;
		if( edgeS == nodeS - 1 )	enS[mnVt] = mxVt;
	}
	for( int i = N - 1; i; -- i )	enS[i] = MIN( enS[i], enS[i + 1] );
	scanf( "%d", &Q );
	while( Q -- > 0 ){
		int queL, queR;
		scanf( "%d%d", &queL, &queR );
		if( enS[queL] > queR )	printf( "Yes\n" );
		else	printf( "No\n" );
	}
	return 0;
}

LOC 2020.11.20 - Prob. 3 / CF396C On Changing Tree

Desc. & Link.

哇哦。

喜闻乐见的 DS 题。

大约是届一下 \(\texttt{lazy tag}\) 的思想。

对于一个节点 \(u\)，只有在路径 \((1,u)\) 上的修改才会产生影响。

这个询问有够简单，差个分即可。

修改就子树加 \(x\)，再加个 \(k\)（除根节点）。

树剖，完了。

#include <cstdio>
#define mod ( 1000000007 )

typedef long long LL;

const int MAXN = 3e5 + 5;

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 &x, _T &y ){ int w = x; x = y; y = w; }

struct starS{
	int to, nx;
	starS( int T = 0, int N = 0 ){ to = T; nx = N; }
} as[MAXN * 2];

struct nodeS{
	int val, add;
	nodeS( int V = 0, int A = 0 ){ val = V; add = A; }
} nodes[MAXN * 4];

int N, M;
int sjc, cnt;
int firS[MAXN], posL[MAXN], posR[MAXN], top[MAXN], son[MAXN], dep[MAXN], fur[MAXN], fa[MAXN];

void pushEdge( const int u, const int v ){ as[++ cnt] = starS( v, firS[u] ); firS[u] = cnt; }

void oneSearch( const int u, const int lst ){
	fa[u] = lst; dep[u] = dep[lst] + 1; fur[u] = 1;
	for( int i = firS[u]; i; i = as[i].nx ){
		int v = as[i].to;
		oneSearch( v, u );
		fur[u] += fur[v];
		if( fur[v] > fur[son[u]] )	son[u] = v;
	}
}

void anotherSearch( const int u, const int nTp ){
	top[u] = nTp; posL[u] = ++ sjc;
	if( son[u] )	anotherSearch( son[u], nTp );
	for( int i = firS[u]; i; i = as[i].nx ){
		int v = as[i].to;
		if( v == son[u] )	continue;
		anotherSearch( v, v );
	}
	posR[u] = sjc;
}

void Spr( const int x, const int l, const int r ){
	if( ! nodes[x].add )	return;
	int mid = ( l + r ) >> 1;
	nodes[x << 1].val = ( nodes[x << 1].val + ( LL )nodes[x].add * ( mid - l + 1 ) % mod ) % mod;
	nodes[x << 1 | 1].val = ( nodes[x << 1 | 1].val + ( LL )nodes[x].add * ( r - mid ) % mod ) % mod;
	nodes[x << 1].add = ( nodes[x << 1].add + nodes[x].add ) % mod;
	nodes[x << 1 | 1].add = ( nodes[x << 1 | 1].add + nodes[x].add ) % mod;
	nodes[x].add = 0;
}

void Upt( const int x ){ nodes[x].val = ( nodes[x << 1].val + nodes[x << 1 | 1].val ) % mod; }

void Modify( const int x, const int l, const int r, const int segL, const int segR, const int segW ){
	if( l > segR || r < segL )	return;
	if( l >= segL && r <= segR ){
		nodes[x].val = ( nodes[x].val + ( LL )segW * ( r - l + 1 ) % mod ) % mod;
		nodes[x].add = ( nodes[x].add + segW ) % mod;
		return;
	}
	int mid = ( l + r ) >> 1;
	Spr( x, l, r );
	Modify( x << 1, l, mid, segL, segR, segW );
	Modify( x << 1 | 1, mid + 1, r, segL, segR, segW );
	Upt( x );
}

int Query( const int x, const int l, const int r, const int segL, const int segR ){
	if( l > segR || r < segL )	return 0;
	if( l >= segL && r <= segR )	return nodes[x].val;
	int mid = ( l + r ) >> 1; Spr( x, l, r );
	return ( Query( x << 1, l, mid, segL, segR ) + Query( x << 1 | 1, mid + 1, r, segL, segR ) ) % mod;
}

int QueryP( int u, int v ){
	int res = 0;
	while( top[u] != top[v] ){
		if( dep[top[u]] < dep[top[v]] )	swapp( u, v );
		res = ( ( res + Query( 1, 1, N, posL[top[u]], posL[u] ) ) % mod + mod ) % mod;
		u = fa[top[u]];
	}
	if( dep[u] >= dep[v] )	swapp( u, v );
	res = ( ( res + Query( 1, 1, N, posL[u], posL[v] ) ) % mod + mod ) % mod;
	return res;
}

int main( ){
	read( N );
	for( int i = 2, lst; i <= N; ++ i ){ read( lst ); pushEdge( lst, i ); }
	oneSearch( 1, 0 ); anotherSearch( 1, 1 );
	read( M );
	while( M -- > 0 ){
		int opt, u, x, k;
		read( opt );
		if( opt == 1 ){
			read( u ); read( x ); read( k );
			Modify( 1, 1, N, posL[u], posL[u], x );
			Modify( 1, 1, N, posL[u] + 1, posR[u], -k );
		}
		else{ read( u ); write( QueryP( 1, u ) ); putchar( '\n' ); }
	}
	return 0;
}

LOC 2020.11.20 - Prob. 4

Desc. & Link.

不想做了。