文章目錄

• A - Arithmetic Sequence
• B - Increasing Triples
• C - 1, 2, 3 - Decomposition
• D - Inc, Dec - Decomposition
• E - Training
• F - Insert Addition

ARC123

A - Arithmetic Sequence

大討論

只能$+1$，先固定中間的數(shù)，看兩邊加誰，如果都是加負(fù)數(shù)說明是中間的數(shù)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

排序一邊，貪心的有盡可能讓小小小，大大大配對出現(xiàn)，雙指針掃即可

#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

結(jié)論：任何一個數(shù)最多不會由超過五個的123類數(shù)組成

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

顯然的定理：對于操作$i$，$B_i=B_{i?1}/C_i=C_{i?1}$二者中至少有一個是不變的

所以，一旦確定了$B_1$，兩個序列都被確定下來，答案也就確定了

反轉(zhuǎn)$C$，重新定義為$A_i=B_i?C_i$，那么$B,C$序列的要求都變成了單調(diào)不下降

最后的答案反正是帶絕對值的，這樣的反轉(zhuǎn)并不影響最后答案的統(tǒng)計，還是${\sum }_{i=1}^n|B_i|+|C_i|$

若$A_{i+1}>A_i$，原問題的解決方案肯定是增加$B$，若$A_{i+1}<A_i$，原問題的解決方案肯定是減少$C$

原則上是沒有變化的，但是改變$C$后，全都變成了增加$B/C$，操作可以表示為

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

對于確定了$B_1$后的序列，每個$B_i,C_i$都是$|B_1?k|$的形式，一共有$2n$個$k$，答案顯然是$B_i$取$k$的中位數(shù)時最小

#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

題目轉(zhuǎn)錄翻譯一下就是求${\sum }_{i=1}^n[A_X+?\frac{i}{B_X}?=A_Y+?\frac{i}{B_Y}?]$

定義$f(x)=A_X+\frac{x}{B_X},g(x)=A_Y+\frac{x}{B_Y},F(x)=?f(x)?,G(x)=?g(x)?$

問題轉(zhuǎn)化為求${\sum }_{i=1}^n[F(i)=G(i)]$

• $f(x)<g(x)?1?F(x)<G(x)$

• $g(x)?1\le f(x)\le g(x)?F(x)=\{G(x)?1,G(x)\}$

  • $g(x)?1\le f(x)$

    $A_Y+\frac{x}{B_Y}?1\le A_X+\frac{x}{B_X}?A_Y{B}_Y{B}_X+x{B}_X?B_X{B}_Y\le A_X{B}_Y{B}_X+x{B}_Y$

    $?(A_Y?A_X?1)B_X{B}_Y\le x(B_Y?B_X)?\frac{(A_Y?A_X?1)B_X{B}_Y}{B_Y?B_X}\le x$

  • $f(x)\le g(x)$

    $\frac{(A_Y?A_X)B_X{B}_Y}{B_Y?B_X}\ge x$

• $g(x)<f(x)\le g(x)+1?F(x)=\{G(x),G(x)+1\}$

  • $g(x)<f(x)$

    $\frac{(A_Y?A_X)B_X{B}_Y}{B_Y?B_X}<x$

  • $f(x)<g(x)+1$

    $\frac{(A_Y?A_X+1)B_X{B}_Y}{B_Y?B_X}>x$

• $g(x)+1<f(x)?F(x)>G(x)$

只有兩種情況才可能出現(xiàn)$F(x)=G(x)$，并且可以數(shù)學(xué)化的解出$f(x)$的范圍

在已知范圍$[l,r]$內(nèi)，可以巧妙轉(zhuǎn)化，${\sum }_{i=l}^r[F(i)=G(i)]?(r?l+1)?|{\sum }_{x=l}^r{A}_X+?\frac{x}{B_X}??{\sum }_{x=l}^r{A}_Y+?\frac{x}{B_Y}?|$

差分地計算${\sum }_{x=l}^r{A}_X+?\frac{x}{B_X}?$

${\sum }_{x=1}^{n}a+?\frac{x}{b}?=b?\frac{(a+a+?\frac{n}{b}??1)??\frac{x}{b}?}{2}+(n??\frac{n}{b}??b+1)?(a+?\frac{x}{b}?)$

#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

數(shù)列的生成類比斯坦納樹

因為$a,b$的范圍，迭代次數(shù)永遠(yuǎn)小于生成數(shù)的大小，因此考慮充分迭代多次后保留$\le n$的數(shù)

如果$(x_1,y_1),(x_2,y_2)$在系數(shù)數(shù)列中相鄰，則有$x_1{y}_2?x_2{y}_1=1$

并且只要坐標(biāo)都$>0$，那么總有一個時刻在系數(shù)數(shù)列中有且僅出現(xiàn)一次

這是個歐幾里得形式的式子，當(dāng)且僅當(dāng)$(a,b)=1$才有解

把數(shù)列當(dāng)成斯坦納樹跑成線段樹，$(i,j)\leftarrow (i?1,j?1)+(i?1,j+1)$

$gcd(a,b)=1?i?a+b?j\le n$，就是莫比烏斯反演了，暴力枚舉$i$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; }

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