zoj 3557 How Many Sets II 求Comb(n,m)%p p很大素数 m<=1e4

本文链接：https://blog.youkuaiyun.com/kongming_acm/article/details/6917039

本文探讨了在给定集合S及其大小n、子集大小m和质数p的情况下，计算满足条件的子集T的数量，并通过模运算进行结果处理。详细介绍了算法实现，包括使用模逆元计算组合数，以及处理不包含连续整数的子集问题。

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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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define bignum long long
bignum exgcd(bignum a,bignum b,bignum& x,bignum& y)
{
if(b==0) return x=1,y=0,a;
bignum r=exgcd(b,a%b,x,y);
bignum t=x;
x=y;
y=t-(a/b)*y;
return r;
}
bignum calc(bignum a,bignum b,bignum mod)
{
bignum x,y;
bignum r=exgcd(b,mod,x,y);
x*=a;
x=(x%mod+mod)%mod;
return x;
}
//C(n,m)%p m<=1e4 p是素数
bignum Comb_p(bignum n,bignum m,bignum p)
{
if(m>n) return 0;//important
if(m>n/2) m=n-m;
bignum a=1,b=1;
bignum nump=0;
for(bignum i=1;i<=m;i++)
{
bignum x=n-i+1,y=i;
while(x%p==0) x/=p,nump++;
while(y%p==0) y/=p,nump--;
a=(a*x)%p;b=(b*y)%p;
}
if(nump) return 0;//important
return calc(a,b,p);
}
//Yi=Xi-Xi-1>=2 Y1>=1 Ym+1>=0 Y1+..Ym+Ym+1=n
int main()
{
int n,m,p;
while(scanf("%d%d%d",&n,&m,&p)==3)
{
int ans=Comb_p((bignum)n-m+1,(bignum)m,(bignum)p);
printf("%d\n",ans);
}
return 0;
}