容斥要么是dp预处理，要么是组合数推导（纯数学式子化简），要么是dfs搜索剪枝

[HAOI2008]硬币购物

problem

solution

先不管$d$的限制，预处理每一个$s$的方案数

然后容斥$-$至少一个超出$d$限制$+$至少两个超出$d$限制$-$至少三个超出$d$限制$+$至少四个超出$d$限制

我们可以强制超出$d$限制，以超出$d_1$为例

强制使用$d_1+1$个$c_1$，那么剩下$s-(d_1+1)c_1$​​的空间随便用即预处理的$d{p}_{s-(d_1+1)c_1}$

code

#include <cstdio>
#define maxn 100005
#define int long long
int tot, c1, c2, c3, c4, d1, d2, d3, d4, s;
int dp[maxn];void init( int c ) {for( int i = c;i <= 1e5;i ++ )dp[i] += dp[i - c];
}signed main() {scanf( "%lld %lld %lld %lld %lld", &c1, &c2, &c3, &c4, &tot );dp[0] = 1;init( c1 ), init( c2 ), init( c3 ), init( c4 );while( tot -- ) {scanf( "%lld %lld %lld %lld %lld", &d1, &d2, &d3, &d4, &s );int ans = dp[s];if( c1 * ( d1 + 1 ) <= s ) ans -= dp[s - c1 * ( d1 + 1 )];if( c2 * ( d2 + 1 ) <= s ) ans -= dp[s - c2 * ( d2 + 1 )];if( c3 * ( d3 + 1 ) <= s ) ans -= dp[s - c3 * ( d3 + 1 )];if( c4 * ( d4 + 1 ) <= s ) ans -= dp[s - c4 * ( d4 + 1 )];if( c1 * ( d1 + 1 ) + c2 * ( d2 + 1 ) <= s ) ans += dp[s - c1 * ( d1 + 1 ) - c2 * ( d2 + 1 )];if( c1 * ( d1 + 1 ) + c3 * ( d3 + 1 ) <= s ) ans += dp[s - c1 * ( d1 + 1 ) - c3 * ( d3 + 1 )];if( c1 * ( d1 + 1 ) + c4 * ( d4 + 1 ) <= s ) ans += dp[s - c1 * ( d1 + 1 ) - c4 * ( d4 + 1 )];if( c2 * ( d2 + 1 ) + c3 * ( d3 + 1 ) <= s ) ans += dp[s - c2 * ( d2 + 1 ) - c3 * ( d3 + 1 )];if( c2 * ( d2 + 1 ) + c4 * ( d4 + 1 ) <= s ) ans += dp[s - c2 * ( d2 + 1 ) - c4 * ( d4 + 1 )];if( c3 * ( d3 + 1 ) + c4 * ( d4 + 1 ) <= s ) ans += dp[s - c3 * ( d3 + 1 ) - c4 * ( d4 + 1 )];if( c1 * ( d1 + 1 ) + c2 * ( d2 + 1 ) + c3 * ( d3 + 1 ) <= s ) ans -= dp[s - c1 * ( d1 + 1 ) - c2 * ( d2 + 1 ) - c3 * ( d3 + 1 )];if( c1 * ( d1 + 1 ) + c2 * ( d2 + 1 ) + c4 * ( d4 + 1 ) <= s ) ans -= dp[s - c1 * ( d1 + 1 ) - c2 * ( d2 + 1 ) - c4 * ( d4 + 1 )];if( c1 * ( d1 + 1 ) + c3 * ( d3 + 1 ) + c4 * ( d4 + 1 ) <= s ) ans -= dp[s - c1 * ( d1 + 1 ) - c3 * ( d3 + 1 ) - c4 * ( d4 + 1 )];if( c2 * ( d2 + 1 ) + c3 * ( d3 + 1 ) + c4 * ( d4 + 1 ) <= s ) ans -= dp[s - c2 * ( d2 + 1 ) - c3 * ( d3 + 1 ) - c4 * ( d4 + 1 )];if( c1 * ( d1 + 1 ) + c2 * ( d2 + 1 ) + c3 * ( d3 + 1 ) + c4 * ( d4 + 1 ) <= s ) ans += dp[s - c1 * ( d1 + 1 ) - c2 * ( d2 + 1 ) - c3 * ( d3 + 1 ) - c4 * ( d4 + 1 )];printf( "%lld\n", ans );}return 0;
}

CF559C Gerald and Giant Chess

problem

solution

到某个点$(i,j)$的方案数为$\left(\genfrac{}{}{0ex}{}{i+j-2}{i-1}\right)$，将终点也看做一个黑格子

设$d{p}_i$表示走到只经过第$i$个黑格子的方案数，枚举可能经过的黑格子$j(x_j\le x_i,y_j\le y_i)$

容斥，$d{p}_i=d{p}_i-{\sum }_{j,x_j\le x_i,y_j\le y_i}d{p}_j\ast \left(\genfrac{}{}{0ex}{}{x_i-x_j+y_i-y_j}{x_i-x_j}\right)$

code

#include <cstdio>
#include <algorithm>
using namespace std;
#define int long long
#define mod 1000000007
#define maxn 200000
struct node {int x, y;node(){}node( int X, int Y ) { x = X, y = Y; }
}pos[maxn];
int fac[maxn + 5], inv[maxn + 5], dp[maxn];
int h, w, n;bool cmp( node MS, node IS ) {return MS.x == IS.x ? MS.y < IS.y : MS.x < IS.x;
}int qkpow( int x, int y ) {int ans = 1;while( y ) {if( y & 1 ) ans = ans * x % mod;x = x * x % mod;y >>= 1;}return ans;
}void init() {fac[0] = inv[0] = 1;for( int i = 1;i <= maxn;i ++ )fac[i] = fac[i - 1] * i % mod;inv[maxn] = qkpow( fac[maxn], mod - 2 );for( int i = maxn - 1;i;i -- )	inv[i] = inv[i + 1] * ( i + 1 ) % mod;
}int C( int n, int m ) {return fac[n] * inv[m] % mod * inv[n - m] % mod;
}signed main() {init();scanf( "%lld %lld %lld", &h, &w, &n );for( int i = 1;i <= n;i ++ )scanf( "%lld %lld", &pos[i].x, &pos[i].y );++ n; pos[n] = node( h, w );sort( pos + 1, pos + n + 1, cmp );for( int i = 1;i <= n;i ++ ) {int x = pos[i].x, y = pos[i].y;dp[i] = C( x + y - 2, x - 1 );for( int j = 1;j < i;j ++ )if( pos[j].x <= x && pos[j].y <= y )dp[i] = ( dp[i] - dp[j] * C( x - pos[j].x + y - pos[j].y, x - pos[j].x ) % mod + mod ) % mod;}printf( "%lld\n", dp[n] );return 0;
}

[SCOI2010]幸运数字

problem

solution

区间$[l,r]$中$i$的倍数个数为$\lfloor \frac{r}{i}\rfloor -\lceil \frac{l}{i}\rceil +1$

容斥，至少选一个幸运数字$-$​至少选两个幸运数字$+$​至少选三个幸运数字$...{\sum }_{i=1}^n(-1{)}^{i-1}$​

如果直接暴搜$1e10$以内所有的幸运数字（$2046$个）的倍数是不可取的

需要搜索剪枝

• 如果幸运数字$i$是幸运数字$j$的倍数，那么$i$就可以不用考虑了
• 如果现在的幸运数字的$lcm$已经超过上限，也果断退出
• 将幸运数字从大到小排序搜索，更容易爆出设定边界，减少前期不必要的分叉搜索

code

#include <cstdio>
#include <algorithm>
using namespace std;
#define ll long long
#define maxn 3000
ll ans, L, R;
int tot, cnt;
ll id[maxn], num[maxn];
bool vis[maxn];void dfs1( int lim, int len, ll now, ll dig ) {if( len > lim ) {id[++ tot] = now;return;}dfs1( lim, len + 1, now + dig * 6, dig * 10 );dfs1( lim, len + 1, now + dig * 8, dig * 10 );
}void init() {for( int i = 1;i <= tot;i ++ ) {if( ! vis[i] ) num[++ cnt] = id[i];for( int j = i + 1;j <= tot;j ++ )if( id[j] % id[i] == 0 ) vis[j] = 1;}
} ll gcd( ll x, ll y ) {if( ! y ) return x;else return gcd( y, x % y );
}ll calc( ll x ) {ll l = L / x + ( L % x != 0 ), r = R / x;return r - l + 1;
}void dfs2( int x, int used, ll lcm ) {if( x > cnt ) {if( ! used ) return;ans += ( used & 1 ? 1 : -1 ) * calc( lcm );return;}dfs2( x + 1, used, lcm );ll t = lcm / gcd( lcm, num[x] );if( 1.0 * t * num[x] <= R )dfs2( x + 1, used + 1, t * num[x] );
}bool cmp( ll x, ll y ) { return x > y; }int main() {scanf( "%lld %lld", &L, &R );for( int i = 1;i <= 10;i ++ ) dfs1( i, 1, 0, 1 );init();sort( num + 1, num + cnt + 1, cmp );dfs2( 1, 0, 1 );printf( "%lld\n", ans );return 0;
}

CF997C Sky Full of Stars

problem

solution

容斥：${\sum }_{i=0}^n{\sum }_{j=0}^n[i+j>0](-1{)}^{i+j+1}{C}_n^i{C}_n^{j}d{p}_{i,j}$

$d{p}_{i,j}:i$行$j$​列颜色相同的方案数

考虑对$d{p}_{i,j}$进行求解，分类讨论

• 只有行/列颜色相同，$i=0/j=0$（以行为例）

  有$i$行颜色相同，每行颜色有$3$种选择，剩下的$n\ast (n-i)$格子是自由选择

  $3^i\times 3^{n\ast (n-i)}$

• 有$i$行$j$列颜色相同

  行与列有交，整个整体只有$3$种选择，剩下的$(n-i)(n-j)$格子是自由选择

  $3\times 3^{(n-i)(n-j)}$

$$d{p}_{i,j}=\{\begin{array}{llc}{3}^i\times 3^{n\ast (n-i)}& & j=0\\ 3^j\times 3^{n\ast (n-j)}& & i=0\\ 3\times 3^{(n-i)(n-j)}& & i\ne 0,j\ne 0\end{array}$$

$i=0/j=0$​的$dp$​​式子比较特殊，考虑单独计算
$$\begin{gathered} \sum _{i=1}^n(-1{)}^{i+1}{C}_n^{i}d{p}_{i,0}+\sum _{j=1}^n(-1{)}^{j+1}{C}_n^{j}d{p}_{j,0} \\ =2\sum _{i=1}^n(-1{)}^{i+1}{C}_n^i{3}^i\cdot 3^{n(n-i)} \end{gathered}$$
然后计算剩下的普通式子
$$\begin{gathered} \sum _{i=1}^n\sum _{j=1}^n(-1{)}^{i+j+1}{C}_n^i{C}_n^j\cdot 3\cdot 3^{(n-i)(n-j)} \\ =\sum _{i=1}^n\sum _{j=1}^n(-1{)}^i(-1{)}^j(-1)C_n^i{C}_n^j{3}^{n^2+1}{3}^{-in}{3}^{-jn}{3}^{ij} \\ =(-1)3^{n^2+1}\sum _{i=1}^n(-1{)}^i{C}_n^i{3}^{-in}\sum _{j=1}^n(-1{)}^j{C}_n^j{3}^{-jn}{3}^{ij} \\ =(-1)3^{n^2+1}\sum _{i=1}^n(-1{)}^i{C}_n^i{3}^{-in}\sum _{j=1}^n{C}_n^j(-1{)}^j{3}^{{(-n+i)}^j} \\ =(-1)3^{n^2+1}\sum _{i=1}^n(-1{)}^i{C}_n^i{3}^{-in}\sum _{j=1}^n{C}_n^j(-3^{-n+i}{)}^j \end{gathered}$$
有二项式定理：$(a+b{)}^n={\sum }_{i=0}^n{C}_n^i{a}^i{b}^{n-i}$

将$a=1,b=-3^{-n+i}$​​带入
$$(a+b{)}^n=(1-3^{-n+i}{)}^n=\sum _{j=0}^n{C}_n^j(-3^{-n+i}{)}^j{1}^{n-j}$$
二项式定理是从$0$的，所以要减去$j=0$的贡献
$$\begin{gathered} (-1)3^{n^2+1}\sum _{i=1}^n(-1{)}^i{C}_n^i{3}^{-in}\sum _{j=1}^n{C}_n^j(-3^{-n+i}{)}^j \\ =(-1)3^{n^2+1}\sum _{i=1}^n((-1{)}^i{C}_n^i{3}^{-in}((1-3^{-n+i}{)}^n-1\left)\right) \end{gathered}$$
两种情况综上，答案为
$$2\sum _{i=1}^n(-1{)}^{i+1}{C}_n^i{3}^i{3}^{n(n-i)}+(-1)3^{n^2+1}\sum _{i=1}^n(-1{)}^i{C}_n^i{3}^{-in}((1-3^{-n+i}{)}^n-1)$$

code

#include <cstdio>
#define int long long
#define mod 998244353
#define maxn 1000005
int n;
int fac[maxn], inv[maxn];int qkpow( int x, int y ) {int ans = 1;while( y ) {if( y & 1 ) ans = ans * x % mod;x = x * x % mod;y >>= 1;}return ans;
}void init() {fac[0] = inv[0] = 1;for( int i = 1;i <= n;i ++ )fac[i] = fac[i - 1] * i % mod;inv[n] = qkpow( fac[n], mod - 2 );for( int i = n - 1;i;i -- )inv[i] = inv[i + 1] * ( i + 1 ) % mod;
}int C( int n, int m ) {return fac[n] * inv[m] % mod * inv[n - m] % mod;
}signed main() {scanf( "%lld", &n );init();int ans = 0, ret = 0;for( int i = 1;i <= n;i ++ )if( i & 1 ) {ans = ( ans - C( n, i ) * qkpow( qkpow( 3, i * n ), mod - 2 ) % mod * ( qkpow( 1 - qkpow( qkpow( 3,  n - i ), mod - 2 ), n ) - 1 ) % mod ) % mod;ret = ( ret + C( n, i ) * qkpow( 3, n * ( n - i ) + i ) ) % mod;}else {ans = ( ans + C( n, i ) * qkpow( qkpow( 3, i * n ), mod - 2 ) % mod * ( qkpow( 1 - qkpow( qkpow( 3,  n - i ), mod - 2 ), n ) - 1 ) % mod ) % mod;ret = ( ret - C( n, i ) * qkpow( 3, n * ( n - i ) + i ) ) % mod;}ans = ans * -1 * qkpow( 3, n * n + 1 ) % mod;ret = ret * 2 % mod;printf( "%lld\n", ( ans + ret + mod + mod ) % mod );return 0;
}

已经没有什么好害怕的了

problem

solution

保证不会有重复的数字暗示了如果有k组糖果比药片能量大那么剩下的n-k组一定都是药片比糖果大

若恰好糖果比药片能量大的组数 比 药片比糖果能量大的组数多K组

反解出$k=\frac{n+K}{2},k$​​表示糖果比药片能量大的组数

先将糖果和药片分别按能量大小排序，设$f_{i,j}$​表示前$i$​颗糖果中有$j$​组满足糖果能量大于药片

$f_{i,j}=f_{i-1,j}+(cn{t}_i-(j-1))f_{i-1,j-1};;cn{t}_i:$​ 药片能量小于糖果$i$​​的药片个数

令$g_i=(n-i)!f_{n,i}$，表示把剩下的$n-i$个随意分配，得到糖果大于药片的组数至少为$i$的方案数

果断标准广义容斥

• $f_i:$ 至多选，$g_i$恰好选

  $f_n={\sum }_{i=0}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)g_i\Rightarrow g_n={\sum }_{i=0}^n(-1{)}^{n-i}(\genfrac{}{}{0ex}{}{n}{i})f_i$​

• $f_i:$ 至少选，$g_i$恰好选

  $f_k={\sum }_{i=k}^n\left(\genfrac{}{}{0ex}{}{i}{k}\right)g_i\Rightarrow g_k={\sum }_{i=k}^n(-1{)}^{i-k}(\genfrac{}{}{0ex}{}{i}{k})f_i$​

一般$f_i$是非常好求的，反演求恰好的$g_i$

$ans={\sum }_{i=k}^n(-1{)}^{i-k}(\genfrac{}{}{0ex}{}{i}{k})g_i$

code

#include <cstdio>
#include <algorithm>
using namespace std;
#define maxn 2005
#define int long long
#define mod 1000000009
int n, k;
int candy[maxn], medicine[maxn], fac[maxn], inv[maxn];
int f[maxn][maxn];int qkpow( int x, int y ) {int ans = 1;while( y ) {if( y & 1 ) ans = ans * x % mod;x = x * x % mod;y >>= 1;}return ans;
}void init() {fac[0] = inv[0] = 1;for( int i = 1;i <= n;i ++ )fac[i] = fac[i - 1] * i % mod;inv[n] = qkpow( fac[n], mod - 2 );for( int i = n - 1;i;i -- )inv[i] = inv[i + 1] * ( i + 1 ) % mod;
}int C( int n, int m ) {return fac[n] * inv[m] % mod * inv[n - m] % mod;
}signed main() {scanf( "%lld %lld", &n, &k );if( ( n + k ) & 1 ) return ! printf( "0\n" );k = ( n + k ) >> 1; init();for( int i = 1;i <= n;i ++ )scanf( "%lld", &candy[i] );for( int i = 1;i <= n;i ++ )scanf( "%lld", &medicine[i] );sort( candy + 1, candy + n + 1 );sort( medicine + 1, medicine + n + 1 );int cnt = 0;f[0][0] = 1; for( int i = 1;i <= n;i ++ ) {f[i][0] = 1;while( cnt < n && medicine[cnt + 1] < candy[i] ) cnt ++;for( int j = 1;j <= i;j ++ )f[i][j] = ( f[i - 1][j] + ( cnt - j + 1 ) * f[i - 1][j - 1] ) % mod;}int ans = 0;for( int i = k;i <= n;i ++ )if( ( i - k ) & 1 )ans = ( ans - C( i, k ) * fac[n - i] % mod * f[n][i] ) % mod;elseans = ( ans + C( i, k ) * fac[n - i] % mod * f[n][i] ) % mod;printf( "%lld\n", ( ans + mod ) % mod );return 0;
}

[JLOI2015]骗我呢

problem

solution

性质：每行必须严格递增，且每行$m$个 ，取值范围被限制于$[0,m]$；所以一行有且只有一个比前面的大$2$​，剩下的一定都是比前面的大恰好$1$

设$d{p}_{i,j}:$ 第$i$行没有出现的数为$j$的方案数

$$d{p}_{i,j}=\sum _{k=1}^{j+1}d{p}_{i-1,k}$$

若第$i$行没有出现$j$，那么$(j,n]$共$n-j$个数都是出现的，除去最大值放在最后一个，也至少需要$n-j-1$个数使得能够满足矩阵要求，所以上一行的上限就是$j+1$不出现，$(j+1,n]$共$n-j-1$个

$$d{p}_{i,j}=\sum _{k=1}^{j+1}d{p}_{i-1,k};d{p}_{i,j-1}=\sum _{k=1}^{j}d{p}_{i-1,k}$$

$$\Rightarrow d{p}_{i,j}=d{p}_{i-1,j+1}+d{p}_{i,j-1}[j>0]$$

转化成二维图形上的关联↓

稍微变换一下（红线是形式上的辅助理解）

对称一下

这完全就是卡特兰数的几何意义解法，非常类似，我们套路的知道可以跟组合数扯上关系

从原点出发，不经过直线$l_1,l_2$，最后到达$(n+m+1,n)$的路径数

$l_1:y=x+1;l_2:y=x-(m+2)$

每一条路径都对应着从原点到翻折点的路径数

每条路径要么先碰$l_1$要么先碰$l_2$，所以用所有方案数减去先碰$l_1$方案数减去先碰$l_2$方案数

减去以$l_1$开头的方案数，需要减去以$l_1,l_1{l}_2$结尾的方案加上$l_2{l}_1,l_2{l}_1{l}_2$结尾的$...$

实现过程：将$(x,y)$沿直线$l_1$​翻折，减去答案，将翻折点沿$l_2$翻折，加上答案$...$

$l_2$开头的同理，$(x,y)\to (y-1,x+1)/(y+m+2,x-m-2)$

code

#include <cstdio>
#include <iostream>
using namespace std;
#define mod 1000000007
#define int long long
#define maxn 5000000
int fac[maxn + 5], inv[maxn + 5];
int n, m;int qkpow( int x, int y ) {int ans = 1;while( y ) {if( y & 1 ) ans = ans * x % mod;x = x * x % mod;y >>= 1;}return ans;
}void init() {fac[0] = inv[0] = 1;for( int i = 1;i <= maxn;i ++ )fac[i] = fac[i - 1] * i % mod;inv[maxn] = qkpow( fac[maxn], mod - 2 );for( int i = maxn - 1;i;i -- )inv[i] = inv[i + 1] * ( i + 1 ) % mod;
}int C( int x, int y ) {if( x < 0 || y < 0 ) return 0;else return fac[x + y] * inv[x] % mod * inv[y] % mod;
}void work1( int &x, int &y ) {swap( x, y ), x --, y ++;
}void work2( int &x, int &y ) {swap( x, y ), x += ( m + 2 ), y -= ( m + 2 );
}signed main() {init();scanf( "%lld %lld", &n, &m );int x = n + m + 1, y = n, ans = C( x, y );while( x >= 0 && y >= 0 ) {work1( x, y );ans = ( ans - C( x, y ) ) % mod;work2( x, y );ans = ( ans + C( x, y ) ) % mod;}x = n + m + 1, y = n;while( x >= 0 && y >= 0 ) {work2( x, y );ans = ( ans - C( x, y ) ) % mod;work1( x, y );ans = ( ans + C( x, y ) ) % mod;}printf( "%lld\n", ( ans + mod ) % mod );return 0;
}