zoj 3557 How Many Sets II

How Many Sets II

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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。

 #include<iostream>
 #include<stdio.h>
 #include<cstring>
 #include<cstdlib>
 using namespace std;
 typedef long long LL;

 LL pow_mod(LL a,LL n,LL p)
 {
     LL ans=;
     while(n)
     {
         if(n&) ans=(ans*a)%p;
         n=n>>;
         a=(a*a)%p;
     }
     return ans;
 }
 LL C(int n,int m,int p)/**Cnm %p **/
 {
     int i;
     LL sum1=,sum2=;
     for(i=;i<=m;i++)
     {
         sum1=(sum1*(n-i+))%p;
         sum2=(sum2*i)%p;
     }
     sum1=(sum1*pow_mod(sum2,p-,p))%p;
     return sum1;
 }
 void solve(int n,int m,int p)
 {
     LL ans=;
     while(n&&m&&ans)
     {
         ans=(ans*C(n%p,m%p,p))%p;
         n=n/p;
         m=m/p;
     }
     printf("%lld\n",ans);
 }
 int main()
 {
     int n,m,p;
     while(scanf("%d%d%d",&n,&m,&p)>)
     {
         n=n-m+;
         if(n<m){
             printf("0\n");
             continue;
         }
         else if(n==m){
             printf("1\n");
             continue;
         }
         if(m>n-m) m=n-m;
         solve(n,m,p);
     }
     return ;
 }