Split Divisibilities (Project Euler 598)
题目大意:
求将$100!$ 拆成$a*b$的方案数,其中$a<=b$并且它们的约数个数一样多。
思路:
先将$100!$质因数分解, 结果如图:
首先想到一个暴力DP, dp[i][j][k]表示考虑完前i个质数, 目前a有j个约数,b有k个约数的方案数。 用map保存状态。
答案就是sum(dp[25][j][j]).
但是状态数会很多(大概有1e8个状态),所以考虑 中途相遇法。 对前3个质数做一次DP, 然后对后面22个质数做一次DP。
最后答案就是 sum (dp1[3][i1][j1] * dp2[22][i2][j2]) 条件是 i1 * i2 = j1 * j2. 即 i1 / j1 = j2 / i2 .
一个优化是只保存 j和k互质的状态。 然后 最后 答案的时候 枚举 i1,j1, 在 dp2中 查找 j2 / i2 = i1 / j1的点 。
因为a<=b,所以最后答案还需要除以2.
代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream>
2 #include <cstdio>
3 #include <algorithm>
4 #include <cmath>
5 #include <set>
6 #include <cstring>
7 #include <map>
8 #include <queue>
9 using namespace std;
10
11 typedef long long ll;
12 #define N 10000000
13 #define M 1100
14 typedef pair<int,int> pii;
15
16
17 bool flag[N];
18 int p[N],phi[N];
19
20 struct node
21 {
22 ll x,y;
23 bool operator < (const node &t)const
24 {
25 return y*t.x<x*t.y;
26 }
27 node (ll _x = 0, ll _y = 0){x = _x; y = _y;}
28 };
29
30 map<node, ll> mp;
31
32 ll Gcd(ll x, ll y)
33 {
34 ll tmp;
35 while (y)
36 {
37 tmp = x % y;
38 x = y, y = tmp;
39 }
40 return x;
41 }
42
43 void Get_Primes(int lim)
44 {
45 phi[1]=1;
46 for (int i=2;i<=lim;i++)
47 {
48 if (!flag[i]) p[++p[0]]=i,phi[i]=i-1;
49 for (int j=1;j<=p[0] && i*p[j]<=lim;j++)
50 {
51 flag[i*p[j]]=true;
52 if (i%p[j]==0)
53 {
54 phi[i*p[j]]=phi[i]*p[j];
55 break;
56 }
57 else phi[i*p[j]]=phi[i]*(p[j]-1);
58 }
59 }
60 }
61
62 map<pair<ll,ll>, ll> f[6], g[23];
63 int cnt[30];
64
65 int main()
66 {
67 freopen("in.in","r",stdin);
68 freopen("out.out","w",stdout);
69
70 int n = 100;
71 Get_Primes(n);
72 for (int i = 1; i <= p[0]; ++i)
73 {
74 int x = p[i];
75 while (x <= n) cnt[i] += n / x, x *= p[i];
76 }
77
78 f[0][make_pair(1,1)] = 1;
79 for (int i = 1; i <= 3; ++i)
80 {
81 for (map<pair<ll,ll>, ll>::iterator it = f[i - 1].begin(); it != f[i - 1].end(); it++)
82 {
83 ll k1 = (*it).first.first, k2 = (*it).first.second;
84 for (int j = 0; j <= cnt[i]; ++j)
85 {
86 ll kx = k1 * (j + 1), ky = k2 * (cnt[i] - j + 1) , d = Gcd(kx, ky);
87 f[i][make_pair(kx / d, ky / d)] += (*it).second;
88 }
89 }
90 }
91 g[0][make_pair(1,1)] = 1;
92 for (int i = 1; i <= p[0] - 3; ++i)
93 {
94 for (map<pair<ll,ll>, ll>::iterator it = g[i - 1].begin(); it != g[i - 1].end(); it++)
95 {
96 ll k1 = (*it).first.first, k2 = (*it).first.second;
97 for (int j = 0; j <= cnt[i + 3]; ++j)
98 {
99 ll kx = k1 * (j + 1), ky = k2 * (cnt[i + 3] - j + 1), d = Gcd(kx, ky);
100 g[i][make_pair(kx / d, ky / d)] += (*it).second;
101 }
102 }
103 }
104 for (map<pair<ll,ll>, ll>::iterator it = g[p[0] - 3].begin(); it != g[p[0] - 3].end(); it++)
105 {
106 ll x = (*it).first.first, y = (*it).first.second;
107 mp[node(x, y)] += (*it).second;
108 }
109
110 ll res = 0;
111 for (map<pair<ll,ll>, ll>::iterator it = f[3].begin(); it != f[3].end(); it++)
112 {
113 ll x = (*it).first.first, y = (*it).first.second;
114 res += (*it).second * mp[node(y, x)];
115 }
116 cout << res / 2 << endl;
117 return 0;
118 }
答案:543194779059
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