「雜項」Nov. 20th, 2020 - Exam. SOL
阿新 • • 發佈：2020-11-21
LOC 2020.11.20 - Prob. 1
\(C=2^{k}\bmod(a+b+c)\)
#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
我先行否決 naive 的線段樹。
輸入暗示連邊？
那就連吧。
考慮每一個連通塊 \(S\)，如果連通塊是個樹，就只能選 \(|S|-1\) 個點。
如果存在環，那麼每個數都取得到。
那麼可以的情況就是詢問區間包含的不是一棵樹。
#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
哇哦。
喜聞樂見的 DS 題。
大約是屆一下 \(\texttt{lazy tag}\) 的思想。
對於一個節點 \(u\)，只有在路徑 \((1,u)\) 上的修改才會產生影響。
這個詢問有夠簡單，差個分即可。
修改就子樹加 \(x\)，再加個 \(k\)（除根節點）。
樹剖，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
不想做了。