A. Contest Start

数学题，分类讨论

一般的，一段区间[l,r][l,r][l,r]会对后面固定人数造成影响，假设是kkk
最后kkk个人，因为自己后面的人数不够kkk，所以贡献总和是k×(k−1)2\frac{k\times(k-1)}{2}2k×(k−1)

显然固定人数为tx\frac{t}{x}xt

#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

前缀和统计[l,r][l,r][l,r]区间内每个字符的个数×\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

贪心。

排个序，把能放在一起的放在一起，然后记录相邻组之间需要的“桥”的个数，用优先队列存储

显然，需要越少越先满足，留下更多的kkk满足后面

#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

贪心模拟，按bbb排序

用最难满足的购买尽可能触碰易满足的bbb

#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

设Li,j:L_{i,j}:Li,j: 左手为第iii个数，右手为第jjj个数，i<ji<ji<j，是否合法

Ri,j:R_{i,j}:Ri,j: 右手为第iii个数，左手为第jjj个数，i<ji<ji<j，是否合法

显然有暴力的n2n^2n2转移
Li,j=[ai∈[al,j,ar,j⋂aj∈[bl,j,br,j]]={Li,j−1i≠j−1∑k=1j−2Rk,j−1i=j−1Ri,j=[ai∈[bl,j,br,j⋂aj∈[al,j,ar,j]]={Ri,j−1i≠j−1∑k=1j−2Lk,j−1i=j−1L_{i,j}=\bigg[a_i\in [a_{l,j},a_{r,j}\bigcap a_j\in[b_{l,j},b_{r,j}]\bigg]=\begin{cases}L_{i,j-1}&&&&&i≠j-1\\\sum_{k=1}^{j-2}R_{k,j-1}&&&&&i=j-1\end{cases}\\R_{i,j}=\bigg[a_i\in [b_{l,j},b_{r,j}\bigcap a_j\in[a_{l,j},a_{r,j}]\bigg]=\begin{cases}R_{i,j-1}&&&&&i≠j-1\\\sum_{k=1}^{j-2}L_{k,j-1}&&&&&i=j-1\end{cases}\\ Li,j=[ai∈[al,j,ar,j⋂aj∈[bl,j,br,j]]={Li,j−1∑k=1j−2Rk,j−1i=j−1i=j−1Ri,j=[ai∈[bl,j,br,j⋂aj∈[al,j,ar,j]]={Ri,j−1∑k=1j−2Lk,j−1i=j−1i=j−1
显然是可以用线段树维护成logloglog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

连着两题线段树好，很好！

定义S1:[l,r]S_1:[l,r]S1:[l,r] 中小于aia_iai的个数，S2:[l,r]S_2:[l,r]S2:[l,r] 中等于aia_iai的个数，S3:[l,r]S_3:[l,r]S3:[l,r] 中大于aia_iai的个数

显然位置iii的奇怪度为
max⁡{S1+S2−⌈S1+S3+S2+12⌉⌈S1+S3+S2+12⌉−(S1+1)\max\begin{cases} S_1+S_2-\lceil\frac{S_1+S_3+S_2+1}{2}\rceil\\ \lceil\frac{S_1+S_3+S_2+1}{2}\rceil-(S_1+1) \end{cases} max{S1+S2−⌈2S1+S3+S2+1⌉⌈2S1+S3+S2+1⌉−(S1+1)
发现：同时在比aia_iai大和比aia_iai小的数中扔掉若干个相同的数，答案并不会发生变化

如此说来有用的只有二者的差，S1S2S3

如果aia_iai放在S2S_2S2的第一位，我们则非常希望max{S3},min{S1}max\{S_3\},min\{S_1\}max{S3},min{S1}这样中位数才会远离S2S_2S2的第一位

如果aia_iai放在S2S_2S2的倒数第一位，我们则非常希望min{S3},max{S1}min\{S_3\},max\{S_1\}min{S3},max{S1}

显然可以线段树维护查询了，将小于等于aia_iai的标记为−1-1−1

对于若干个相同的aia_iai位置，先查再改是一种情况，改了再查是另一种情况

#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 ++ ) {//放在众多相同的最后一个->后面的尽量少前面的尽量多 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对[i,n]造成的-1影响 然后再变成1后的影响 所以一次性加2 for( int j = 0;j < s[i].size();j ++ ) {//放在众多相同的第一个->后面的尽量多前面的尽量少 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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