A - Arithmetic Sequence

大讨论

只能 $+ 1$ ，先固定中间的数，看两边加谁，如果都是加负数说明是中间的数需要加

#include <cstdio>
#include <iostream>
using namespace std;
#define int long long
int x, y, z;int Fabs( int x ) {return x < 0 ? -x : x;
}signed main() {scanf( "%lld %lld %lld", &x, &y, &z );if( x > z ) swap( x, z );int d1 = y - x, d2 = z - y;int x1 = y - d2, z1 = y + d1;if( x1 >= x ) printf( "%lld\n", x1 - x );else if( z1 >= z ) printf( "%lld\n", z1 - z );else if( ( z - x ) & 1 ) printf( "%lld\n", x + ( z - x ) / 2 - y + 2 );else printf( "%lld\n", x + ( z - x ) / 2 - y );return 0;
}

B - Increasing Triples

排序一边，贪心的有尽可能让小小小，大大大配对出现，双指针扫即可

#include <cstdio>
#include <algorithm>
using namespace std;
#define maxn 100005
int A[maxn], B[maxn], C[maxn];
int n;int main() {scanf( "%d", &n );for( int i = 1;i <= n;i ++ )scanf( "%d", &A[i] );for( int i = 1;i <= n;i ++ )scanf( "%d", &B[i] );for( int i = 1;i <= n;i ++ )scanf( "%d", &C[i] );sort( A + 1, A + n + 1 );sort( B + 1, B + n + 1 );sort( C + 1, C + n + 1 );int ipB = 1, ipC = 1, ans = 0;for( int i = 1;i <= n;i ++ ) {while( ipB <= n && B[ipB] <= A[i] ) ipB ++;while( ipC <= n && C[ipC] <= B[ipB] ) ipC ++;if( ipB > n || ipC > n ) break;else ans ++, ipB ++, ipC ++;}printf( "%d\n", ans );return 0;
}

C - 1, 2, 3 - Decomposition

结论：任何一个数最多不会由超过五个的123类数组成

暴力算每各位check，1/2/3/4/5依次是否可以

#include <map>
#include <cstdio>
using namespace std;
#define int long long
map < int, int > mp;
int T, n;int solve( int n ) {if( ! n ) return 0;if( mp[n] ) return mp[n];int s = n / 10, r = n % 10;if( 1 <= r && r <= 3 && solve( s ) <= 1 ) return mp[n] = 1;if( 2 <= r && r <= 6 && solve( s ) <= 2 ) return mp[n] = 2;if( 3 <= r && r <= 9 && solve( s ) <= 3 ) return mp[n] = 3;if( 4 <= r && r <= 9 && solve( s ) <= 4 ) return mp[n] = 4;if( 0 <= r && r <= 2 && solve( s - 1 ) <= 4 ) return mp[n] = 4;return mp[n] = 5;
}signed main() {scanf( "%lld", &T );while( T -- ) {scanf( "%lld", &n );printf( "%lld\n", solve( n ) );	}return 0;
}

D - Inc, Dec - Decomposition

显然的定理：对于操作 $i$ ， $B_i=B_{i-1}/C_i=C_{i-1}$ 二者中至少有一个是不变的

所以，一旦确定了 $B_1$ ，两个序列都被确定下来，答案也就确定了

反转 $C$ ，重新定义为 $A_i=B_i-C_i$ ，那么 $B, C$ 序列的要求都变成了单调不下降

最后的答案反正是带绝对值的，这样的反转并不影响最后答案的统计，还是 $∑i=1n∣Bi∣+∣Ci∣\sum_{i=1}^n|B_i|+|C_i|$

若 $A_{i+1}>A_i$ ，原问题的解决方案肯定是增加 $B$ ，若 $A_{i+1}<A_i$ ，原问题的解决方案肯定是减少 $C$

原则上是没有变化的，但是改变 $C$ 后，全都变成了增加 $B / C$ ，操作可以表示为

$B_{i+1}=B_{i}+\max(A_{i+1}-A_i,0)$
$C_{i+1}=C_i+\max(A_i-A_{i+1},0)$

对于确定了 $B_1$ 后的序列，每个 $B_i,C_i$ 都是 $B_1-k|$ 的形式，一共有 $2 n$ 个 $k$ ，答案显然是 $B_i$ 取 $k$ 的中位数时最小

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
#define int long long
#define maxn 200005
vector < int > ans;
int n;
int A[maxn];int Fabs( int x ) {return x < 0 ? -x : x;
}signed main() {scanf( "%lld", &n );for( int i = 1;i <= n;i ++ )scanf( "%lld", &A[i] );int B = A[1], C = 0;ans.push_back( B ), ans.push_back( C );for( int i = 1;i < n;i ++ ) {int x = A[i + 1] - A[i];if( x >= 0 ) B += x;else C -= x;ans.push_back( B );ans.push_back( C );}sort( ans.begin(), ans.end() );int ret = 0;for( int i = 0;i < ans.size();i ++ )ret += Fabs( ans[i] - ans[n] );printf( "%lld\n", ret );return 0;
}

E - Training

题目转录翻译一下就是求 $∑i=1n[AX+⌊iBX⌋=AY+⌊iBY⌋]\sum_{i=1}^n\bigg[A_X+\lfloor\frac{i}{B_X}\rfloor=A_Y+\lfloor\frac{i}{B_Y}\rfloor\bigg]$

定义 $f(x)=AX+xBX,g(x)=AY+xBY,F(x)=⌊f(x)⌋,G(x)=⌊g(x)⌋f(x)=A_X+\frac{x}{B_X},g(x)=A_Y+\frac{x}{B_Y},F(x)=\lfloor f(x)\rfloor,G(x)=\lfloor g(x)\rfloor$

问题转化为求 $∑i=1n[F(i)=G(i)]\sum_{i=1}^n\bigg[F(i)=G(i)\bigg]$

$f(x)<g(x)−1⇒F(x)<G(x)f(x)<g(x)-1\Rightarrow F(x)<G(x)$
$g(x)−1≤f(x)≤g(x)⇒F(x)={G(x)−1,G(x)}g(x)-1\le f(x)\le g(x)\Rightarrow F(x)=\bigg\{G(x)-1,G(x)\bigg\}$
- $g(x)−1≤f(x)g(x)-1\le f(x)$
  
  $AY+xBY−1≤AX+xBX⇔AYBYBX+xBX−BXBY≤AXBYBX+xBYA_Y+\frac{x}{B_Y}-1\le A_X+\frac{x}{B_X}\Leftrightarrow A_YB_YB_X+xB_X-B_XB_Y\le A_XB_YB_X+xB_Y$
  
  $⇔(AY−AX−1)BXBY≤x(BY−BX)⇔(AY−AX−1)BXBYBY−BX≤x\Leftrightarrow (A_Y-A_X-1)B_XB_Y\le x(B_Y-B_X)\Leftrightarrow \frac{(A_Y-A_X-1)B_XB_Y}{B_Y-B_X}\le x$
- $f(x)≤g(x)f(x)\le g(x)$
  
  $(AY−AX)BXBYBY−BX≥x\frac{(A_Y-A_X)B_XB_Y}{B_Y-B_X}\ge x$
$g(x)<f(x)≤g(x)+1⇒F(x)={G(x),G(x)+1}g(x)<f(x)\le g(x)+1\Rightarrow F(x)=\bigg\{G(x),G(x)+1\bigg\}$
- $g (x) < f (x)$
  
  $(AY−AX)BXBYBY−BX<x\frac{(A_Y-A_X)B_XB_Y}{B_Y-B_X}< x$
- $f (x) < g (x) + 1$
  
  $(AY−AX+1)BXBYBY−BX>x\frac{(A_Y-A_X+1)B_XB_Y}{B_Y-B_X}> x$
$g(x)+1<f(x)⇒F(x)>G(x)g(x)+1<f(x)\Rightarrow F(x)>G(x)$

只有两种情况才可能出现 $F (x) = G (x)$ ，并且可以数学化的解出 $f (x)$ 的范围

在已知范围 $[l, r]$ 内，可以巧妙转化， $∑i=lr[F(i)=G(i)]⇔(r−l+1)−∣∑x=lrAX+⌊xBX⌋−∑x=lrAY+⌊xBY⌋∣\sum_{i=l}^r\bigg[F(i)=G(i)\bigg]\Leftrightarrow (r-l+1)-\bigg|\sum_{x=l}^rA_X+\lfloor\frac{x}{B_X}\rfloor-\sum_{x=l}^{r}A_Y+\lfloor\frac{x}{B_Y}\rfloor\bigg|$

差分地计算 $∑x=lrAX+⌊xBX⌋\sum_{x=l}^rA_X+\lfloor\frac{x}{B_X}\rfloor$

$∑x=1na+⌊xb⌋=b∗(a+a+⌊nb⌋−1)∗⌊xb⌋2+(n−⌊nb⌋∗b+1)∗(a+⌊xb⌋)\sum_{x=1}^na+\lfloor\frac{x}{b}\rfloor=b*\frac{(a+a+\lfloor\frac{n}{b}\rfloor-1)*\lfloor\frac{x}{b}\rfloor}{2}+(n-\lfloor\frac{n}{b}\rfloor*b+1)*(a+\lfloor\frac{x}{b}\rfloor)$

#include <cmath>
#include <cstdio>
#include <iostream>
using namespace std;
#define int long long
#define eps 1e-8int calc( int a, int b, int n ) {int t = ( n + 1 ) / b;return ( a * 2 + ( t - 1 ) ) * t * b / 2 + ( a + t ) * ( n - t * b + 1 );
}signed main() {int T, n, AX, AY, BX, BY;scanf( "%lld", &T );long double L, R; int l, r, ans, cnt;while( T -- ) {scanf( "%lld %lld %lld %lld %lld", &n, &AX, &BX, &AY, &BY );if( BX == BY ) {if( AX == AY ) printf( "%lld\n", n );else printf( "0\n" );continue;}if( BX > BY ) swap( AX, AY ), swap( BX, BY );ans = 0;//g(x)-1<=f(x)<=g(x)L = 1.0 * ( AY - AX - 1 ) * BX * BY / ( BY - BX ) + eps;R = ( AY - AX ) * 1.0 * BX * BY / ( BY - BX );if( L < 1 ) l = 1;else l = ceil( L );if( R > n ) r = n;else r = floor( R );r = max( r, l - 1 );cnt = calc( AY, BY, r ) - calc( AY, BY, l - 1 ) - calc( AX, BX, r ) + calc( AX, BX, l - 1 );ans += r - l + 1 - cnt;//g(x)<f(x)<g(x)+1L = R + eps;R = ( AY - AX + 1 ) * 1.0 * BX * BY / ( BY - BX ) - eps;if( L < 1 ) l = 1;else l = ceil( L );if( R > n ) r = n;else r = floor( R );r = max( r, l - 1 );cnt = calc( AX, BX, r ) - calc( AX, BX, l - 1 ) - calc( AY, BY, r ) + calc( AY, BY, l - 1 );ans += r - l + 1 - cnt;printf( "%lld\n", ans );} return 0;
}

F - Insert Addition

数列的生成类比斯坦纳树

因为 $a, b$ 的范围，迭代次数永远小于生成数的大小，因此考虑充分迭代多次后保留 $≤n\le n$ 的数

如果 $x_1,y_1),(x_2,y_2)$ 在系数数列中相邻，则有 $x_1y_2-x_2y_1=1$

并且只要坐标都 $> 0$ ，那么总有一个时刻在系数数列中有且仅出现一次

这是个欧几里得形式的式子，当且仅当 $(a, b) = 1$ 才有解

把数列当成斯坦纳树跑成线段树， $(i,j)←(i−1,j−1)+(i−1,j+1)(i,j)\leftarrow (i-1,j-1)+(i-1,j+1)$

$gcd(a,b)=1⇒i∗a+b∗j≤ngcd(a,b)=1\Rightarrow i*a+b*j\le n$ ，就是莫比乌斯反演了，暴力枚举 $i$ 暴力算

#include <cstdio>
#define int long long
#define maxn 300005
int mu[maxn], prime[maxn];
bool vis[maxn];
int a, b, n, l, r;void init() {int cnt = 0;mu[1] = 1;for( int i = 2;i <= n;i ++ ) {if( ! vis[i] ) prime[++ cnt] = i, mu[i] = -1;for( int j = 1;j <= cnt && i * prime[j] <= n;j ++ ) {vis[i * prime[j]] = 1;if( i % prime[j] == 0 ) break;else mu[i * prime[j]] = -mu[i];}}
}void print( int x ) {if( l > r || x > n ) return;else if( l == 1 ) printf( "%lld ", x ), r --;else l --, r --;
}int calc( int x, int y ) {int cnt = 0;for( int i = 1;x + y <= n / i;i ++ ) {if( ! mu[i] ) continue;for( int j = 1;x * j + y <= n / i;j ++ )cnt += mu[i] * ( n / i - x * j ) / y;}return cnt;
}void print_all( int x, int y ) {if( x + y > n ) return;print_all( x, x + y );print( x + y );print_all( x + y, y );
}void dfs( int x, int y ) {int cnt = calc( x, y );if( l > r || ! cnt ) return;if( l > cnt ) {l -= cnt, r -= cnt;return;}if( l == 1 && cnt <= r ) {print_all( x, y );return;}dfs( x, x + y );print( x + y );dfs( x + y, y );
}signed main() {scanf( "%lld %lld %lld %lld %lld", &a, &b, &n, &l, &r );init();print( a ), dfs( a, b ), print( b );return 0;
}