HDOJ4373 Mysterious For

Posted 2020-06-14

1. 题目描述
有两种不同类型的循环，并给出一个由1、2组成的序列，表示嵌套的循环类型。
问这样组着的循环一共需要多少次循环？并将结果模364875103。

2.基本思路
显然，每当遇到一个类型1的序列，即可以判定12...2的嵌套循环共多少次，而1类型的循环次数为常亮。
因此，将原序列从1分开，并将每个子序列的循环次数相乘即为总的循环次数。
1     共循环n次 = C[n][1]
12   共循环n*(n+1)/2次 = C[n+1][2]
122 共循环n*(n+1)*(n+2)/6次 = C[n+2][3]
12..2 |2|=m 共循环n*(n+1)*(n+2)/6次 = C[n+m-1][m]
由Lucas定理可知，假设p为素数，有

比较悲剧的是364875103是个合数，将它拆解为两个素数并使用LUCAS后，
我们可以知道ans=a1(mod p1), ans=a2(mod p2), 且p1, p2互素，求ans = ? (mod p1*p2)。
使用中国剩余定理。
使用欧拉定理求逆。

3. 代码

  1 /* 4373 */
  2 #include <iostream>
  3 #include <sstream>
  4 #include <string>
  5 #include <map>
  6 #include <queue>
  7 #include <set>
  8 #include <stack>
  9 #include <vector>
 10 #include <deque>
 11 #include <algorithm>
 12 #include <cstdio>
 13 #include <cmath>
 14 #include <ctime>
 15 #include <cstring>
 16 #include <climits>
 17 #include <cctype>
 18 #include <cassert>
 19 #include <functional>
 20 #include <iterator>
 21 #include <iomanip>
 22 using namespace std;
 23 //#pragma comment(linker,"/STACK:102400000,1024000")
 24 
 25 #define sti                set<int>
 26 #define stpii            set<pair<int, int> >
 27 #define mpii            map<int,int>
 28 #define vi                vector<int>
 29 #define pii                pair<int,int>
 30 #define vpii            vector<pair<int,int> >
 31 #define rep(i, a, n)     for (int i=a;i<n;++i)
 32 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 33 #define clr                clear
 34 #define pb                 push_back
 35 #define mp                 make_pair
 36 #define fir                first
 37 #define sec                second
 38 #define all(x)             (x).begin(),(x).end()
 39 #define SZ(x)             ((int)(x).size())
 40 #define lson            l, mid, rt<<1
 41 #define rson            mid+1, r, rt<<1|1
 42 #define LL                __int64
 43 
 44 const int maxn = 25;
 45 const int mod1 = 97;
 46 const int mod2 = 3761599;
 47 const int mod = 364875103;
 48 LL fact1[mod1+15];
 49 LL fact2[mod2+15];
 50 LL e1, e2;
 51 int pos[maxn];
 52 int n, m, k;
 53 
 54 LL Pow(LL base, LL n, LL mod) {
 55     LL ret = 1;
 56     
 57     base %= mod;
 58     while (n) {
 59         if (n & 1)
 60             ret = ret * base % mod;
 61         base = base * base % mod;
 62         n >>= 1;
 63     }
 64     
 65     return ret;
 66 }
 67 
 68 void init() {
 69     fact1[0] = fact2[0] = 1;
 70     rep(i, 1, mod1)
 71         fact1[i] = fact1[i-1] * i % mod1;
 72     rep(i, 1, mod2)
 73         fact2[i] = fact2[i-1] * i % mod2;
 74     e1 = mod2 * Pow(mod2, mod1-2, mod1);
 75     e2 = mod1 * Pow(mod1, mod2-2, mod2);
 76     
 77     #ifndef ONLINE_JUDGE
 78         printf("e1 = %I64d, e2 = %I64d\n", e1, e2);
 79     #endif
 80 }
 81 
 82 LL C(LL n, LL m, LL mod, LL *fact) {
 83     if (n < m)    return 0;
 84     return fact[n] * Pow(fact[m]*fact[n-m], mod-2, mod) % mod;
 85 }
 86 
 87 LL Lucas(LL n, LL m, LL mod, LL *fact) {
 88     if (m == 0)    return 1;
 89     return C(n%mod, m%mod, mod, fact) * Lucas(n/mod, m/mod, mod, fact);
 90 }
 91 
 92 void solve() {
 93     LL ans = 1, tmp;
 94     LL a1, a2;
 95     
 96     rep(i, 0, k) {
 97         m = pos[i+1]-pos[i];
 98         a1 = Lucas(n+m-1, m, mod1, fact1);
 99         a2 = Lucas(n+m-1, m, mod2, fact2);
100         tmp = (a1*e1 + a2*e2) % mod;
101         #ifndef ONLINE_JUDGE
102             printf("a1 = %I64d, a2=%I64d, tmp=%I64d\n", a1, a2, tmp);
103         #endif
104         ans = ans * tmp % mod;
105     }
106     
107     printf("%I64d\n", ans);
108 }
109 
110 int main() {
111     ios::sync_with_stdio(false);
112     #ifndef ONLINE_JUDGE
113         freopen("data.in", "r", stdin);
114         freopen("data.out", "w", stdout);
115     #endif
116     
117     int t;
118     
119     init();
120     scanf("%d", &t);
121     rep(tt, 1, t+1) {
122         scanf("%d%d%d", &n, &m, &k);
123         rep(i, 0, k)
124             scanf("%d", &pos[i]);
125         pos[k] = m;
126         printf("Case #%d: ", tt);
127         solve();
128     }
129     
130     #ifndef ONLINE_JUDGE
131         printf("time = %d.\n", (int)clock());
132     #endif
133     
134     return 0;
135 }

4. 数据生成器

 1 from copy import deepcopy
 2 from random import randint, shuffle
 3 import shutil
 4 import string
 5 
 6 
 7 def GenDataIn():
 8     with open("data.in", "w") as fout:
 9         t = 20
10         bound = 10**5
11         fout.write("%d\n" % (t))
12         for tt in xrange(t):
13             n = randint(20, bound)
14             m = randint(20, bound)
15             k = randint(1, 15)
16             fout.write("%d %d\n%d " % (n, m, k))
17             L = [0]
18             st = set()
19             for i in xrange(1, k):
20                 while True:
21                     x = randint(1, m)
22                     if x not in st:
23                         break
24                 L.append(x)
25                 st.add(x)
26             L = sorted(L)
27             fout.write(" ".join(map(str, L)) + "\n")
28             
29             
30 def MovDataIn():
31     desFileName = "F:\eclipse_prj\workspace\hdoj\data.in"
32     shutil.copyfile("data.in", desFileName)
33 
34     
35 if __name__ == "__main__":
36     GenDataIn()
37     MovDataIn()