[NowCoder5673E]Enigmatic Partition

题意

定义整数$n$的拆分$n=a_1+a_2+...+a_m$为"enigmatic partition"为符合以下条件的拆分：

1. $a_i\in Z$且$1\le a_i\le n$。
2. $a_i\le a_{i+1}\le a_i+1$
3. $a_m=a_1+2$

设$f(n)$表示$n$有多少种不同的"enigmatic partition"，求$\sum _{i=l}^{r}f(i)$.

$1\le l\le r\le 10^5$

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写在前面

考场上对$m$分类打表归纳出了$f(n)$的数学表达式，但是不知道如何用数学证明，如果有大佬知道怎么证明请联系一下我，谢谢。

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题解

首先可以写一个爆搜

#include
using namespace std;
const int N=100+5;
int n,a[N],f[N];
void dfs(int m,int res){
	if(res<0)return;
	if(res==0){
		--m;
		if(a[m]==a[1]+2)++f[n];
		return;
	}
	a[m]=a[m-1];
	dfs(m+1,res-a[m]);
	if(a[m-1]<=a[1]+1){
		a[m]=a[m-1]+1;
		dfs(m+1,res-a[m]);
	}
}
int main(){
	for(n=1;n<=50;++n){
		for(int i=1;i<=n;++i)
			a[1]=i,
			dfs(2,n-i);
	}
	for(n=1;n<=50;++n)
		printf("%2d:%3d\n",n,f[n]);
    return 0;
}

$n$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$	$13$	$14$	$15$	$16$	$17$	$18$	$19$	$20$	$21$	$22$	$23$	$24$	$25$	$26$	$27$	$28$	$29$	$30$	$31$	$32$	$33$	$34$	$35$	$36$	$37$	$38$	$39$	$40$	$41$	$42$	$43$	$44$	$45$	$46$	$47$	$48$	$49$	$50$
$f(n)$	$0$	$0$	$0$	$0$	$0$	$1$	$1$	$2$	$4$	$4$	$6$	$9$	$10$	$11$	$17$	$17$	$21$	$24$	$28$	$31$	$38$	$37$	$45$	$50$	$56$	$56$	$68$	$69$	$78$	$83$	$91$	$94$	$107$	$106$	$122$	$126$	$136$	$137$	$155$	$158$	$171$	$176$	$190$	$193$	$214$	$211$	$231$	$238$	$254$	$256$

粗略地看只有$f(2k+1)$和$f(2k+2)$相差较小，但和$f(2k)$相差较大的规律。

通过简易分析，可知当$m=n-2,n-3$且$n\ge 8$时一定只有一种方案；当$m=n-4$且$n\ge 9$时只有两种方案。

考虑对不同的$m$进行分类打表：

int g[N][N];
void dfs(int m,int res){
	...
	if(res==0){
		--m;
		if(a[m]==a[1]+2)
			++g[n][m];
		return;
	}
	...
}
int main(){
	...
	for(n=1;n<=50;++n){
		printf("%2d:",n);
		for(int m=1;m<=n;++m)
			printf("%3d ",g[n][m]);
		puts("");
	}
	...
}

得到如下表格

	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20	21	22	23	24	25	26	27	28	29	30	31	32	33	34	35	36	37	38	39	40	41	42	43	44	45	46	47	48	49	50
1	$0$
2	$0$	$0$
3	$0$	$0$	$0$
4	$0$	$0$	$0$	$0$
5	$0$	$0$	$0$	$0$	$0$
6	$0$	$0$	$1$	$0$	$0$	$0$
7	$0$	$0$	$0$	$1$	$0$	$0$	$0$
8	$0$	$0$	$0$	$1$	$1$	$0$	$0$	$0$
9	$0$	$0$	$1$	$1$	$1$	$1$	$0$	$0$	$0$
10	$0$	$0$	$0$	$0$	$2$	$1$	$1$	$0$	$0$	$0$
11	$0$	$0$	$0$	$1$	$1$	$2$	$1$	$1$	$0$	$0$	$0$
12	$0$	$0$	$1$	$1$	$1$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
13	$0$	$0$	$0$	$1$	$1$	$2$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
14	$0$	$0$	$0$	$0$	$1$	$1$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
15	$0$	$0$	$1$	$1$	$2$	$2$	$2$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
16	$0$	$0$	$0$	$1$	$1$	$1$	$2$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
17	$0$	$0$	$0$	$1$	$1$	$2$	$2$	$3$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
18	$0$	$0$	$1$	$0$	$1$	$2$	$2$	$2$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
19	$0$	$0$	$0$	$1$	$1$	$2$	$2$	$3$	$3$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
20	$0$	$0$	$0$	$1$	$2$	$1$	$2$	$2$	$3$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
21	$0$	$0$	$1$	$1$	$1$	$2$	$3$	$3$	$3$	$4$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
22	$0$	$0$	$0$	$0$	$1$	$1$	$2$	$2$	$3$	$3$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
23	$0$	$0$	$0$	$1$	$1$	$2$	$2$	$3$	$3$	$4$	$4$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
24	$0$	$0$	$1$	$1$	$1$	$2$	$2$	$3$	$3$	$3$	$4$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
25	$0$	$0$	$0$	$1$	$2$	$2$	$2$	$3$	$3$	$4$	$4$	$5$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
26	$0$	$0$	$0$	$0$	$1$	$1$	$2$	$2$	$3$	$3$	$4$	$4$	$6$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
27	$0$	$0$	$1$	$1$	$1$	$2$	$2$	$3$	$4$	$4$	$4$	$5$	$5$	$6$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
28	$0$	$0$	$0$	$1$	$1$	$1$	$3$	$2$	$3$	$3$	$4$	$4$	$5$	$6$	$6$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
29	$0$	$0$	$0$	$1$	$1$	$2$	$2$	$3$	$3$	$4$	$4$	$5$	$5$	$6$	$6$	$6$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
30	$0$	$0$	$1$	$0$	$2$	$2$	$2$	$2$	$3$	$4$	$4$	$4$	$5$	$5$	$7$	$6$	$6$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
31	$0$	$0$	$0$	$1$	$1$	$2$	$2$	$3$	$3$	$4$	$4$	$5$	$5$	$6$	$6$	$7$	$6$	$6$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
32	$0$	$0$	$0$	$1$	$1$	$1$	$2$	$3$	$3$	$3$	$4$	$4$	$5$	$5$	$6$	$7$	$7$	$6$	$6$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
33	$0$	$0$	$1$	$1$	$1$	$2$	$2$	$3$	$3$	$4$	$5$	$5$	$5$	$6$	$6$	$7$	$7$	$7$	$6$	$6$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
34	$0$	$0$	$0$	$0$	$1$	$1$	$2$	$2$	$3$	$3$	$4$	$4$	$5$	$5$	$6$	$6$	$8$	$7$	$7$	$6$	$6$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
35	$0$	$0$	$0$	$1$	$2$	$2$	$3$	$3$	$3$	$4$	$4$	$5$	$5$	$6$	$6$	$7$	$7$	$8$	$7$	$7$	$6$	$6$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
36	$0$	$0$	$1$	$1$	$1$	$2$	$2$	$2$	$4$	$3$	$4$	$5$	$5$	$5$	$6$	$6$	$7$	$8$	$8$	$7$	$7$	$6$	$6$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
37	$0$	$0$	$0$	$1$	$1$	$2$	$2$	$3$	$3$	$4$	$4$	$5$	$5$	$6$	$6$	$7$	$7$	$8$	$8$	$8$	$7$	$7$	$6$	$6$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
38	$0$	$0$	$0$	$0$	$1$	$1$	$2$	$2$	$3$	$3$	$4$	$4$	$5$	$5$	$6$	$6$	$7$	$7$	$9$	$8$	$8$	$7$	$7$	$6$	$6$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
39	$0$	$0$	$1$	$1$	$1$	$2$	$2$	$3$	$3$	$4$	$4$	$5$	$6$	$6$	$6$	$7$	$7$	$8$	$8$	$9$	$8$	$8$	$7$	$7$	$6$	$6$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
40	$0$	$0$	$0$	$1$	$2$	$1$	$2$	$3$	$3$	$4$	$4$	$4$	$5$	$5$	$6$	$6$	$7$	$7$	$8$	$9$	$9$	$8$	$8$	$7$	$7$	$6$	$6$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
41	$0$	$0$	$0$	$1$	$1$	$2$	$2$	$3$	$3$	$4$	$4$	$5$	$5$	$6$	$6$	$7$	$7$	$8$	$8$	$9$	$9$	$9$	$8$	$8$	$7$	$7$	$6$	$6$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
42	$0$	$0$	$1$	$0$	$1$	$2$	$3$	$2$	$3$	$3$	$4$	$4$	$5$	$6$	$6$	$6$	$7$	$7$	$8$	$8$	$10$	$9$	$9$	$8$	$8$	$7$	$7$	$6$	$6$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
43	$0$	$0$	$0$	$1$	$1$	$2$	$2$	$3$	$3$	$4$	$4$	$5$	$5$	$6$	$6$	$7$	$7$	$8$	$8$	$9$	$9$	$10$	$9$	$9$	$8$	$8$	$7$	$7$	$6$	$6$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
44	$0$	$0$	$0$	$1$	$1$	$1$	$2$	$2$	$3$	$3$	$5$	$4$	$5$	$5$	$6$	$6$	$7$	$7$	$8$	$8$	$9$	$10$	$10$	$9$	$9$	$8$	$8$	$7$	$7$	$6$	$6$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
45	$0$	$0$	$1$	$1$	$2$	$2$	$2$	$3$	$4$	$4$	$4$	$5$	$5$	$6$	$7$	$7$	$7$	$8$	$8$	$9$	$9$	$10$	$10$	$10$	$9$	$9$	$8$	$8$	$7$	$7$	$6$	$6$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
46	$0$	$0$	$0$	$0$	$1$	$1$	$2$	$2$	$3$	$3$	$4$	$4$	$5$	$5$	$6$	$6$	$7$	$7$	$8$	$8$	$9$	$9$	$11$	$10$	$10$	$9$	$9$	$8$	$8$	$7$	$7$	$6$	$6$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
47	$0$	$0$	$0$	$1$	$1$	$2$	$2$	$3$	$3$	$4$	$4$	$5$	$5$	$6$	$6$	$7$	$7$	$8$	$8$	$9$	$9$	$10$	$10$	$11$	$10$	$10$	$9$	$9$	$8$	$8$	$7$	$7$	$6$	$6$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
48	$0$	$0$	$1$	$1$	$1$	$2$	$2$	$3$	$3$	$3$	$4$	$5$	$5$	$5$	$6$	$7$	$7$	$7$	$8$	$8$	$9$	$9$	$10$	$11$	$11$	$10$	$10$	$9$	$9$	$8$	$8$	$7$	$7$	$6$	$6$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
49	$0$	$0$	$0$	$1$	$1$	$2$	$3$	$3$	$3$	$4$	$4$	$5$	$5$	$6$	$6$	$7$	$7$	$8$	$8$	$9$	$9$	$10$	$10$	$11$	$11$	$11$	$10$	$10$	$9$	$9$	$8$	$8$	$7$	$7$	$6$	$6$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
50	$0$	$0$	$0$	$0$	$2$	$1$	$2$	$2$	$3$	$4$	$4$	$4$	$5$	$5$	$6$	$6$	$7$	$7$	$8$	$8$	$9$	$9$	$10$	$10$	$12$	$11$	$11$	$10$	$10$	$9$	$9$	$8$	$8$	$7$	$7$	$6$	$6$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$

可以看出非常明显的规律，特别的当$m\ge \frac{n}{2}$时，$g[n][m]$的值为确定的常数，且每到偶数行就会多出一个常数，归纳后结果为$\lfloor \frac{n-2}{4}\rfloor$。

当$m<\frac{n}{2}$时，继续观察可以发现：

1. $m=2k+3$时，这一列几乎全是$k$，会出现少数$k+1$。
2. $m=2k+4$时，这一列几乎全是$k$和$k+1$交替出现，少量$k$会变成$k+1$

事出反常必有妖，观察这些反常的$k+1$出现的规律，可以总结得到

1. $m=2k+3$时，反常的$k+1$出现在$2k+3$的倍数行，即$m$的倍数行。
2. $m=2k+4$时，反常的$k+1$出现在$2k+4$的倍数行，即$m$的倍数行。

大胆猜测，$f(n)$的值与其约数个数$d(n)$有关。

任取$1$行，对比其将反常的数修正前后的$g[n][m]$，得到：

$42$	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20	21	22	23	24	25	26	27	28	29	30	31	32	33	34	35	36	37	38	39	40	41	42
修正前	$0$	$0$	$1$	$0$	$1$	$2$	$3$	$2$	$3$	$3$	$4$	$4$	$5$	$6$	$6$	$6$	$7$	$7$	$8$	$8$	$10$	$9$	$9$	$8$	$8$	$7$	$7$	$6$	$6$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$
修正后	$0$	$0$	$0$	$0$	$1$	$1$	$2$	$2$	$3$	$3$	$4$	$4$	$5$	$5$	$6$	$6$	$7$	$7$	$8$	$8$	$10$	$9$	$9$	$8$	$8$	$7$	$7$	$6$	$6$	$5$	$5$	$4$	$4$	$3$	$3$	$2$	$2$	$1$	$1$	$0$	$0$	$0$

可以发现所有$n$的大于$2$，小于$\frac{n}{2}$的约数列都出现了反常的$k+1$。

总结一下前面的规律可以得到
$$f(n)-f(n-1)=\text{两行反常的k+1出现次数差}+\text{常数}$$

而每行反常的$k+1$出现次数又与$d(n)$有关，通过进一步观察发现次数差恰好等于$d(n)-d(n-1)$。

回到最初的表格，算出$h(n)=f(n)-f(n-1)-[d(n)-d(n-1)]$

$n$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$	$13$	$14$	$15$	$16$	$17$	$18$	$19$	$20$	$21$	$22$	$23$	$24$	$25$	$26$	$27$	$28$	$29$	$30$	$31$	$32$	$33$	$34$	$35$	$36$	$37$	$38$	$39$	$40$	$41$	$42$	$43$	$44$	$45$	$46$	$47$	$48$	$49$	$50$
$h(n)$	$-1$	$-1$	$0$	$-1$	$1$	$-1$	$2$	$-1$	$3$	$-1$	$4$	$-1$	$5$	$-1$	$6$	$-1$	$7$	$-1$	$8$	$-1$	$9$	$-1$	$10$	$-1$	$11$	$-1$	$12$	$-1$	$13$	$-1$	$14$	$-1$	$15$	$-1$	$16$	$-1$	$17$	$-1$	$18$	$-1$	$19$	$-1$	$20$	$-1$	$21$	$-1$	$22$	$-1$	$23$	$-1$

归纳得
$$h(n)=\{\begin{array}{rlr}& -1,& 2\mid n\\ & \frac{n-3}{2},& 2\nmid n\end{array}$$

用累加法可以求得

$$f(n)-f(1)-[d(n)-d(1)]=\sum _{i=2}^{n}h(i)=\{\begin{array}{rlr}& -2k+\frac{(k-1)(k-2)}{2},& n=2k+1\\ & -2k-1+\frac{(k-1)(k-2)}{2},& n=2k+2\end{array}$$
其中$k=\lfloor \frac{n-1}{2}\rfloor$，化简可得

$$f(n)=d(n)+\frac{1}{2}\lfloor \frac{n-1}{2}{\rfloor }^2-\frac{3}{2}\lfloor \frac{n-1}{2}\rfloor -\frac{1}{2}[3+(-1{)}^n]$$

记$s(n)=\sum _{i=1}^{n}f(i)$，令$k=\lfloor \frac{n-1}{2}\rfloor$，当$n=2k+1$时，有
$$\begin{array}{rl}s(n)& =f(2k+1)+\sum _{i=0}^{k-1}[f(2i+1)+f(2i+2)]\\ & =\sum _{i=1}^{n}d(i)+\frac{1}{2}(k^2-3k-2)+\sum _{i=0}^{k-1}(i^2-3i-3)\\ & =\sum _{i=1}^{n}d(i)+\frac{1}{2}(k^2-3k-2)+\sum _{i=1}^{k-1}{i}^2-3\sum _{i=1}^{k-1}i-3k\end{array}$$

同理，当$n=2k+2$时
$$s(n)=\sum _{i=1}^{n}d(i)+\sum _{i=1}^k{i}^2-3\sum _{i=1}^{k}i-3(k+1)$$

其中$\sum _{i=1}^{n}d(i)=\sum _{i=1}^n\lfloor \frac{n}{i}\rfloor$可以根号分块处理，$\sum _{i=1}^k{i}^2$和$\sum _{i=1}^{k}i$可以用求和公式$O(1)$处理，故单次询问的复杂度为$O(\sqrt{n})$。

算法一：线性筛+递推

时间复杂度$O(n+T)$

#include 
using namespace std;
const int N = 1e5 + 5;
typedef long long ll;
int is[N], pr[N], d[N], c[N];
ll f[N], s[N];
int main() {
    int n = 1e5;
    d[1] = 1;
    for (int i = 2; i <= n; ++i) {
        if (!is[i])
            pr[++pr[0]] = i, d[i] = 2, c[i] = 1;
        for (int j = 1, x; j <= pr[0] && (x = i * pr[j]) <= n; ++j) {
            is[x] = 1;
            if (i % pr[j])
                c[x] = 1, d[x] = d[i] << 1;
            else {
                c[x] = c[i] + 1, d[x] = d[i] / c[x] * (c[x] + 1);
                break;
            }
        }
    }
    for (int i = 1; i <= n; ++i)
        f[i] = f[i - 1] + d[i] - d[i - 1] + (i & 1 ? (i - 3) / 2 : -1),
        s[i] = s[i - 1] + f[i];
    scanf("%*d");
    for (int i = 1, l, r; ~scanf("%d%d", &l, &r); ++i)
        printf("Case #%d: %lld\n", i, s[r] - s[l - 1]);
    return 0;
}

算法二：根号分块

时间复杂度$O(T\sqrt{n})$

#include 
using namespace std;
typedef long long ll;
inline ll s1(int n) { return (ll)n * (n + 1) >> 1; }
inline ll s2(int n) { return (ll)n * (n + 1) * (n << 1 | 1) / 6; }
inline ll s(int n) {
    if (!n)
        return 0;
    int k = (n - 1) / 2, f = n & 1;
    ll Sum = n & 1 ? ((ll)k * (k - 3) - 2) >> 1 : 0ll;
    for (int i = 1, j; i <= n; i = j + 1) {
        j = n / (n / i);
        Sum += (ll)(n / i) * (j - i + 1);
    }
    Sum += s2(k - f) - 3 * s1(k - f) - 3 * (k + 1 - f);
    return Sum;
}
int main() {
    scanf("%*d");
    for (int i = 1, l, r; ~scanf("%d%d", &l, &r); ++i)
        printf("Case #%d: %lld\n", i, s(r) - s(l - 1));
    return 0;
}

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后记

赛后发现，把此题扩展到$a_m=a_1+k$，对$m$分类打出来的表找依旧很有规律，稍微总结一下应该也能做。