文章目錄

• A. Contest Start
• B. Love Song
• C. Stable Groups
• D. PriceFixed
• E. Game with Cards
• F. Strange Array

#727-Div.2

A. Contest Start

數學題，分類討論

• 一般的，一段區間$[l,r]$會對后面固定人數造成影響，假設是$k$
• 最后$k$個人，因為自己后面的人數不夠$k$，所以貢獻總和是$\frac{k\times (k?1)}{2}$

顯然固定人數為$\frac{t}{x}$

#include <cstdio> #define int long long int T, n, x, t;signed main() {scanf( "%lld", &T );while( T -- ) {scanf( "%lld %lld %lld", &n, &x, &t );int k = t / x;if( n < k ) printf( "%lld\n", n * ( n - 1 ) / 2 );else printf( "%lld\n", ( n - k ) * k + ( k - 1 ) * k / 2 );}return 0; }

B. Love Song

前綴和統計$[l,r]$區間內每個字符的個數$\times$字符在字母表的排位

#include <cstdio> #define maxn 100005 int n, Q; char s[maxn]; int cnt[maxn][26];int main() {scanf( "%d %d %s", &n, &Q, s + 1 );for( int i = 1;i <= n;i ++ ) {for( int j = 0;j < 26;j ++ )cnt[i][j] = cnt[i - 1][j];cnt[i][s[i] - 'a'] ++;}while( Q -- ) {int l, r;scanf( "%d %d", &l, &r );int ans = 0;for( int i = 0;i < 26;i ++ )ans += ( i + 1 ) * ( cnt[r][i] - cnt[l - 1][i] );printf( "%d\n", ans );}return 0; }

C. Stable Groups

貪心。

排個序，把能放在一起的放在一起，然后記錄相鄰組之間需要的“橋”的個數，用優先隊列存儲

顯然，需要越少越先滿足，留下更多的$k$滿足后面

#include <queue> #include <cstdio> #include <algorithm> using namespace std; #define int long long #define maxn 200005 priority_queue < int, vector < int >, greater < int > > q; int n, k, x; int a[maxn];signed main() {scanf( "%lld %lld %lld", &n, &k, &x );for( int i = 1;i <= n;i ++ )scanf( "%lld", &a[i] );sort( a + 1, a + n + 1 );int cnt = 1;for( int i = 1;i < n;i ++ )if( a[i + 1] - a[i] > x ) {cnt ++;q.push( ( a[i + 1] - a[i] - 1 ) / x );}while( ! q.empty() ) {if( q.top() <= k ) k -= q.top(), cnt --, q.pop();else break;}printf( "%lld\n", cnt );return 0; }

D. PriceFixed

貪心模擬，按$b$排序

用最難滿足的購買盡可能觸碰易滿足的$b$

#include <cstdio> #include <algorithm> using namespace std; #define int long long #define maxn 100005 struct node {int a, b; }it[maxn]; int n;bool cmp( node x, node y ) {return x.b > y.b; }signed main() {scanf( "%lld", &n );for( int i = 1;i <= n;i ++ )scanf( "%lld %lld", &it[i].a, &it[i].b );sort( it + 1, it + n + 1, cmp );int tot = 0, ans = 0;for( int i = 1;i <= n;i ++ ) {if( tot >= it[n].b ) {ans += it[n].a;tot += it[n].a;it[n].a = 0;i --;n --;}else if( tot + it[i].a < it[n].b ) {tot += it[i].a;ans += ( it[i].a << 1 );it[i].a = 0;} else {int t = it[n].b - tot;it[i].a -= t;ans += ( t << 1 );tot += t;ans += it[n].a;tot += it[n].a;it[n].a = 0;n --;i --;}}if( ! it[n].a )printf( "%lld\n", ans );else {if( tot >= it[n].b )printf( "%lld\n", ans + it[n].a );else if( it[n].a + tot <= it[n].b )printf( "%lld\n", ans + ( it[n].a << 1 ) );else {int t = it[n].b - tot;ans += ( t << 1 );it[n].a -= t;ans += it[n].a;printf( "%lld\n", ans );}}return 0; }

E. Game with Cards

設$L_{i,j}:$ 左手為第$i$個數，右手為第$j$個數，$i<j$，是否合法

$R_{i,j}:$ 右手為第$i$個數，左手為第$j$個數，$i<j$，是否合法

顯然有暴力的$n^2$轉移
$$\begin{gathered} L_{i,j}=[a_i\in [a_{l,j},a_{r,j}?a_j\in [b_{l,j},b_{r,j}]]=\{\begin{array}{llcccc}{L}_{i,j?1}& & & & & i=j?1\\ {\sum }_{k=1}^{j?2}{R}_{k,j?1}& & & & & i=j?1\end{array} \\ {R}_{i,j}=[a_i\in [b_{l,j},b_{r,j}?a_j\in [a_{l,j},a_{r,j}]]=\{\begin{array}{llcccc}{R}_{i,j?1}& & & & & i=j?1\\ {\sum }_{k=1}^{j?2}{L}_{k,j?1}& & & & & i=j?1\end{array} \end{gathered}$$
顯然是可以用線段樹維護成$log$的

#include <cstdio> #define maxn 100005 int n, m;struct SegMentTree {struct node {int l, r, tag, id, s;}t[maxn * 30];int cnt = 0, root;#define lson t[now].l#define rson t[now].rvoid pushdown( int now ) {if( t[now].tag ) {t[lson].s = t[rson].s = 0;t[lson].tag = t[rson].tag = t[now].tag;t[now].tag = 0;} }void clear( int now, int l, int r, int L, int R ) {if( r < L || R < l ) return;if( ! now ) return;if( L <= l && r <= R ) {t[now].s = 0;t[now].tag = 1;return;}pushdown( now );int mid = ( l + r ) >> 1;clear( lson, l, mid, L, R );clear( rson, mid + 1, r, L, R );t[now].s = t[lson].s | t[rson].s;}void insert( int &now, int l, int r, int pos, int id ) {if( ! now ) now = ++ cnt;if( l == r ) {t[now].s = 1;t[now].id = id;return;}pushdown( now );int mid = ( l + r ) >> 1;if( pos <= mid ) insert( lson, l, mid, pos, id );else insert( rson, mid + 1, r, pos, id );t[now].s = t[lson].s | t[rson].s;}int query( int now = 1, int l = 0, int r = m ) {if( l == r ) return t[now].id;pushdown( now );int mid = ( l + r ) >> 1;if( t[lson].s ) return query( lson, l, mid );else return query( rson, mid + 1, r );}bool check() { return t[1].s; }}L, R;struct read {int x, L_rangel, L_ranger, R_rangel, R_ranger;read(){}read( int X, int AL, int AR, int BL, int BR ) {x = X, L_rangel = AL, L_ranger = AR, R_rangel = BL, R_ranger = BR;} }MS[maxn];int ansl[maxn], ansr[maxn], ans[maxn]; bool ChooseL[maxn], ChooseR[maxn];int main() {scanf( "%d %d", &n, &m );L.insert( L.root, 0, m, 0, 0 );R.insert( R.root, 0, m, 0, 0 );for( int i = 1, x, L_rangel, L_ranger, R_rangel, R_ranger;i <= n;i ++ ) {scanf( "%d %d %d %d %d", &x, &L_rangel, &L_ranger, &R_rangel, &R_ranger );MS[i] = read( x, L_rangel, L_ranger, R_rangel, R_ranger );bool l_ok = L.check(), r_ok = R.check();if( l_ok ) ansl[i] = L.query();else ansl[i] = -1;if( r_ok ) ansr[i] = R.query();else ansr[i] = -1;if( L_rangel <= x && x <= L_ranger ) {//x is included in L->[l,r] which is suitable for request of option 0R.clear( R.root, 0, m, 0, R_rangel - 1 );R.clear( R.root, 0, m, R_ranger + 1, m );if( i > 1 && R_rangel <= MS[i - 1].x && MS[i - 1].x <= R_ranger && l_ok )R.insert( R.root, 0, m, MS[i - 1].x, i - 1 ), ChooseR[i] = 1;//however we must also sure that the number we keep on right hand is also included in R->[l,r] //only both are acceptable can we do option 0 truly}else R.clear( R.root, 0, m, 0, m );//otherwise the whole array of L should be clearedif( R_rangel <= x && x <= R_ranger ) {L.clear( L.root, 0, m, 0, L_rangel - 1 );L.clear( L.root, 0, m, L_ranger + 1, m );if( i > 1 && L_rangel <= MS[i - 1].x && MS[i - 1].x <= L_ranger && r_ok )L.insert( L.root, 0, m, MS[i - 1].x, i - 1 ), ChooseL[i] = 1;}else L.clear( L.root, 0, m, 0, m );}if( L.check() || R.check() ) {//backward print(END -> BEGIN)printf( "Yes\n" );int pos = n, lim, f;if( L.check() ) lim = L.query(), f = 0;else lim = R.query(), f = 1;while( pos ) {if( f ) {//this option choose 0(replace left)ans[pos] = 0;if( pos > lim + 1 ) goto out;else f ^= 1, lim = ansl[pos];}else {//this option choose 1(replace right)ans[pos] = 1;if( pos > lim + 1 ) goto out;else f ^= 1, lim = ansr[pos];}out : pos --;}for( int i = 1;i <= n;i ++ )printf( "%d ", ans[i] );}else printf( "No\n" );return 0; }

F. Strange Array

連著兩題線段樹好，很好！

定義$S_1:[l,r]$ 中小于$a_i$的個數，$S_2:[l,r]$ 中等于$a_i$的個數，$S_3:[l,r]$ 中大于$a_i$的個數

顯然位置$i$的奇怪度為
$$\max \{\begin{array}{l}{S}_1+S_2??\frac{S_1+S_3+S_2+1}{2}?\\ ?\frac{S_1+S_3+S_2+1}{2}??(S_1+1)\end{array}$$
發現：同時在比$a_i$大和比$a_i$小的數中扔掉若干個相同的數，答案并不會發生變化

如此說來有用的只有二者的差，S1S2S3

如果$a_i$放在$S_2$的第一位，我們則非常希望$max\{S_3\},min\{S_1\}$這樣中位數才會遠離$S_2$的第一位

如果$a_i$放在$S_2$的倒數第一位，我們則非常希望$min\{S_3\},max\{S_1\}$

顯然可以線段樹維護查詢了，將小于等于$a_i$的標記為$?1$

對于若干個相同的$a_i$位置，先查再改是一種情況，改了再查是另一種情況

#include <cstdio> #include <vector> #include <iostream> #include <algorithm> using namespace std; #define maxn 200005 struct node {int maxx, minn, lazy; }t[maxn << 2]; vector < int > s[maxn]; int n; int ans[maxn];void pushdown( int num ) {t[num << 1].maxx += t[num].lazy;t[num << 1].minn += t[num].lazy;t[num << 1].lazy += t[num].lazy;t[num << 1 | 1].maxx += t[num].lazy;t[num << 1 | 1].minn += t[num].lazy;t[num << 1 | 1].lazy += t[num].lazy;t[num].lazy = 0; }void modify( int num, int l, int r, int L, int R, int val ) {if( L <= l && r <= R ) {t[num].lazy += val;t[num].maxx += val;t[num].minn += val;return;}pushdown( num );int mid = ( l + r ) >> 1;if( L <= mid ) modify( num << 1, l, mid, L, R, val );if( mid < R ) modify( num << 1 | 1, mid + 1, r, L, R, val );t[num].maxx = max( t[num << 1].maxx, t[num << 1 | 1].maxx );t[num].minn = min( t[num << 1].minn, t[num << 1 | 1].minn ); }int query_max( int num, int l, int r, int L, int R ) {if( L > R ) return 0; if( R < l || r < L ) return -1e9;if( L <= l && r <= R ) return t[num].maxx;pushdown( num );int mid = ( l + r ) >> 1;return max( query_max( num << 1, l, mid, L, R ), query_max( num << 1 | 1, mid + 1, r, L, R ) ); }int query_min( int num, int l, int r, int L, int R ) {if( L > R ) return 0;if( R < l || r < L ) return 1e9;if( L <= l && r <= R ) return t[num].minn;pushdown( num );int mid = ( l + r ) >> 1;return min( query_min( num << 1, l, mid, L, R ), query_min( num << 1 | 1, mid + 1, r, L, R ) ); }int main() {scanf( "%d", &n );for( int i = 1, x;i <= n;i ++ ) {scanf( "%d", &x );s[x].push_back( i );modify( 1, 1, n, i, n, -1 );}for( int i = 1;i <= n;i ++ ) {for( int j = 0;j < s[i].size();j ++ ) {//放在眾多相同的最后一個->后面的盡量少前面的盡量多 int k = s[i][j];int l = max( 0, query_max( 1, 1, n, 1, k - 1 ) );//compare with 0 means just choosing k itselfint r = query_min( 1, 1, n, k, n );ans[k] = max( ans[k], ( l - r + 2 ) / 2 - 1 );}for( int j = 0;j < s[i].size();j ++ )modify( 1, 1, n, s[i][j], n, 2 );//抵消掉之前i對[i,n]造成的-1影響 然后再變成1后的影響 所以一次性加2 for( int j = 0;j < s[i].size();j ++ ) {//放在眾多相同的第一個->后面的盡量多前面的盡量少 int k = s[i][j];int l = min( 0, query_min( 1, 1, n, 1, k - 1 ) );int r = query_max( 1, 1, n, k, n );ans[k] = max( ans[k], ( r - l ) - ( r - l + 2 ) / 2 );}}for( int i = 1;i <= n;i ++ )printf( "%d ", ans[i] );return 0; }

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