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
/* 4373 */
#include <iostream>
#include <sstream>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#include <deque>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <climits>
#include <cctype>
#include <cassert>
#include <functional>
#include <iterator>
#include <iomanip>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,1024000") #define sti set<int>
#define stpii set<pair<int, int> >
#define mpii map<int,int>
#define vi vector<int>
#define pii pair<int,int>
#define vpii vector<pair<int,int> >
#define rep(i, a, n) for (int i=a;i<n;++i)
#define per(i, a, n) for (int i=n-1;i>=a;--i)
#define clr clear
#define pb push_back
#define mp make_pair
#define fir first
#define sec second
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define lson l, mid, rt<<1
#define rson mid+1, r, rt<<1|1
#define LL __int64 const int maxn = ;
const int mod1 = ;
const int mod2 = ;
const int mod = ;
LL fact1[mod1+];
LL fact2[mod2+];
LL e1, e2;
int pos[maxn];
int n, m, k; LL Pow(LL base, LL n, LL mod) {
LL ret = ; base %= mod;
while (n) {
if (n & )
ret = ret * base % mod;
base = base * base % mod;
n >>= ;
} return ret;
} void init() {
fact1[] = fact2[] = ;
rep(i, , mod1)
fact1[i] = fact1[i-] * i % mod1;
rep(i, , mod2)
fact2[i] = fact2[i-] * i % mod2;
e1 = mod2 * Pow(mod2, mod1-, mod1);
e2 = mod1 * Pow(mod1, mod2-, mod2); #ifndef ONLINE_JUDGE
printf("e1 = %I64d, e2 = %I64d\n", e1, e2);
#endif
} LL C(LL n, LL m, LL mod, LL *fact) {
if (n < m) return ;
return fact[n] * Pow(fact[m]*fact[n-m], mod-, mod) % mod;
} LL Lucas(LL n, LL m, LL mod, LL *fact) {
if (m == ) return ;
return C(n%mod, m%mod, mod, fact) * Lucas(n/mod, m/mod, mod, fact);
} void solve() {
LL ans = , tmp;
LL a1, a2; rep(i, , k) {
m = pos[i+]-pos[i];
a1 = Lucas(n+m-, m, mod1, fact1);
a2 = Lucas(n+m-, m, mod2, fact2);
tmp = (a1*e1 + a2*e2) % mod;
#ifndef ONLINE_JUDGE
printf("a1 = %I64d, a2=%I64d, tmp=%I64d\n", a1, a2, tmp);
#endif
ans = ans * tmp % mod;
} printf("%I64d\n", ans);
} int main() {
ios::sync_with_stdio(false);
#ifndef ONLINE_JUDGE
freopen("data.in", "r", stdin);
freopen("data.out", "w", stdout);
#endif int t; init();
scanf("%d", &t);
rep(tt, , t+) {
scanf("%d%d%d", &n, &m, &k);
rep(i, , k)
scanf("%d", &pos[i]);
pos[k] = m;
printf("Case #%d: ", tt);
solve();
} #ifndef ONLINE_JUDGE
printf("time = %d.\n", (int)clock());
#endif return ;
}
4. 数据生成器
from copy import deepcopy
from random import randint, shuffle
import shutil
import string def GenDataIn():
with open("data.in", "w") as fout:
t = 20
bound = 10**5
fout.write("%d\n" % (t))
for tt in xrange(t):
n = randint(20, bound)
m = randint(20, bound)
k = randint(1, 15)
fout.write("%d %d\n%d " % (n, m, k))
L = [0]
st = set()
for i in xrange(1, k):
while True:
x = randint(1, m)
if x not in st:
break
L.append(x)
st.add(x)
L = sorted(L)
fout.write(" ".join(map(str, L)) + "\n") def MovDataIn():
desFileName = "F:\eclipse_prj\workspace\hdoj\data.in"
shutil.copyfile("data.in", desFileName) if __name__ == "__main__":
GenDataIn()
MovDataIn()