数 sqrt 缩小范围
整除分块
1 #include <cstdio>
2 #include <cstdlib>
3 #include <cmath>
4 #include <cstring>
5 #include <string>
6 #include <algorithm>
7 #include <iostream>
8 using namespace std;
9 #define ll long long
10
11 const int maxn=1e5+10;
12 const ll mod=1e9+7;
13 const double eps=1e-8;
14
15 ll f[110][maxn],add[maxn],cnt[maxn];
16
17 /**
18 大于sqrt(maxvalue)的x，
19 肯定是其它数到x，到从x到其它数
20
21 计数 用 整除分块
22 **/
23
24 int main()
25 {
26 ll n,siz,mid,mmid,l,r,i,j,g=0,sum=0;
27 scanf("%lld%lld",&siz,&n);
28 mmid=sqrt(siz+eps);
29 mid=siz/(mmid+1);
30 g=0;
31 for (l=1;l<=siz;l=r+1)
32 {
33 ///[l,r]
34 r=siz/(siz/l);
35 // printf("%lld %lld %lld\n",l,r,siz/l);
36 if (siz/l<=mid)
37 cnt[siz/l]=r-l+1;
38 }
39
40 for (j=1;j<=mmid;j++)
41 f[1][j]=1;
42 for (i=2;i<=n;i++)
43 {
44 g=0;
45 for (j=1;j<=mmid;j++)
46 (g+=f[i-1][j])%=mod;
47 for (j=1;j<=mmid;j++)
48 f[i][j]=g;
49
50 for (l=1;l<=mid;l++)
51 add[l]=(add[l-1]+f[i-2][l])%mod;
52
53 g=0;
54 for (l=mid;l>=1;l--)
55 {
56 (g+=add[l]*cnt[l])%mod;
57 (f[i][l]+=g)%=mod;
58 }
59 }
60
61
62 for (j=1;j<=mmid;j++)
63 (sum+=f[n][j])%mod;
64 for (l=1;l<=mid;l++)
65 {
66 add[l]=(add[l-1]+f[n-1][l])%mod;
67 (sum+=add[l]*cnt[l])%=mod;
68 }
69
70 ///
71 memset(f,0,sizeof(f));
72 g=0;
73 for (l=mid;l>=1;l--)
74 {
75 (g+=cnt[l])%mod;
76 f[1][l]=g;
77 }
78 n--;
79 for (i=2;i<=n;i++)
80 {
81 g=0;
82 for (j=1;j<=mmid;j++)
83 (g+=f[i-1][j])%=mod;
84 for (j=1;j<=mmid;j++)
85 f[i][j]=g;
86
87 for (l=1;l<=mid;l++)
88 add[l]=(add[l-1]+f[i-2][l])%mod;
89
90 g=0;
91 for (l=mid;l>=1;l--)
92 {
93 (g+=add[l]*cnt[l])%mod;
94 (f[i][l]+=g)%=mod;
95 }
96 }
97
98 for (j=1;j<=mmid;j++)
99 (sum+=f[n][j])%mod;
100 for (l=1;l<=mid;l++)
101 {
102 add[l]=(add[l-1]+f[n-1][l])%mod;
103 (sum+=add[l]*cnt[l])%=mod;
104 }
105
106
107 printf("%lld",sum);
108 return 0;
109 }
110 /*
111 special
112 100=10*10
113
114 100 3
115 1 1 100
116 2 2 50
117 3 3 33
118 4 4 25
119 5 5 20
120 6 6 16
121 7 7 14
122 8 8 12
123 9 9 11
124 10 10 10
125 ///
126 11 11 9
127 12 12 8
128 13 14 7
129 15 16 6
130 17 20 5
131 21 25 4
132 26 33 3
133 34 50 2
134 51 100 1
135
136
137 23 3
138 1 1 23
139 2 2 11
140 3 3 7
141 4 4 5
142 ///
143 5 5 4
144 6 7 3
145 8 11 2
146 12 23 1
147
148
149 31622*log(n) *100
150
151 10 5
152 1 1 10
153 2 2 5
154 3 3 3
155
156 4 5 2
157 6 10 1
158
159 */
来源：https://www.cnblogs.com/cmyg/p/11108256.html