zoj 3557 How Many Sets II

How Many Sets II
Time Limit: 2 Seconds      Memory Limit: 65536 KB

Given a set S = {1, 2, ..., n}, number m and p, your job is to count how many set T satisfies the following condition:

  • T is a subset of S
  • |T| = m
  • T does not contain continuous numbers, that is to say x and x+1 can not both in T

 

Input

There are multiple cases, each contains 3 integers n ( 1 <= n <= 109 ), m ( 0 <= m <= 104, m <= n ) and p ( p is prime, 1 <= p <= 109 ) in one line seperated by a single space, proceed to the end of file.

Output

Output the total number mod p.

Sample Input

5 1 11
5 2 11

Sample Output

5
6

 由于m<=10^4,所以只需要一个for循环计算sum1,sum2就可以了。

 和fzu 2020是一样的。

 方案数目为C(n-k+1,k)种。

 1 #include<iostream>
 2 #include<stdio.h>
 3 #include<cstring>
 4 #include<cstdlib>
 5 using namespace std;
 6 typedef long long LL;
 7 
 8 LL pow_mod(LL a,LL n,LL p)
 9 {
10     LL ans=1;
11     while(n)
12     {
13         if(n&1) ans=(ans*a)%p;
14         n=n>>1;
15         a=(a*a)%p;
16     }
17     return ans;
18 }
19 LL C(int n,int m,int p)/**Cnm %p **/
20 {
21     int i;
22     LL sum1=1,sum2=1;
23     for(i=1;i<=m;i++)
24     {
25         sum1=(sum1*(n-i+1))%p;
26         sum2=(sum2*i)%p;
27     }
28     sum1=(sum1*pow_mod(sum2,p-2,p))%p;
29     return sum1;
30 }
31 void solve(int n,int m,int p)
32 {
33     LL ans=1;
34     while(n&&m&&ans)
35     {
36         ans=(ans*C(n%p,m%p,p))%p;
37         n=n/p;
38         m=m/p;
39     }
40     printf("%lld\n",ans);
41 }
42 int main()
43 {
44     int n,m,p;
45     while(scanf("%d%d%d",&n,&m,&p)>0)
46     {
47         n=n-m+1;
48         if(n<m){
49             printf("0\n");
50             continue;
51         }
52         else if(n==m){
53             printf("1\n");
54             continue;
55         }
56         if(m>n-m) m=n-m;
57         solve(n,m,p);
58     }
59     return 0;
60 }

 

posted @ 2014-05-10 11:18  芷水  阅读(231)  评论(0)  编辑  收藏  举报