luogu1313计算系数题解--二项式定理

Posted 2021-01-06 rye-catcher

题目链接

https://www.luogu.org/problemnew/show/P1313

分析

二项式定理

((a+b)^n=sum_{k=0}^{n}{C^k_n a^k b^{n-k} })

于是我们要求的即是(C^k_n imes a^n imes b^m),于是直接快速幂，然后按公式(C^k_n=frac {n!}{(n-k)! imes k!}),化成(prod_{i=k+1}^{i<=n} i imes ((n-k)!)^{-1})

代码

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <algorithm>
#include <cctype>
#define ll long long 
#define ri register int 
using std::min;
using std::max;
template <class T>inline void read(T &x){
    x=0;int ne=0;char c;
    while(!isdigit(c=getchar()))ne=c==‘-‘;
    x=c-48;
    while(isdigit(c=getchar()))x=(x<<3)+(x<<1)+c-48;
    x=ne?-x:x;return ;
}
const ll p=10007;
int n,m,k,a,b;
inline ll ksm(ll a,ll c){
    ll ans=1;
    while(c){
        if(c&1)ans=ans*a%p;
        a=a*a%p;
        c=c>>1;
    }
    return ans;
}
ll fermat(ll a){
    return ksm(a,p-2);
}
int main(){
    /*ans=a^n*b^m*C(n,k)*/
    ll ans=1;
    read(a),read(b),read(k),read(n),read(m);
    ans=ksm(a,n)*ksm(b,m)%p;
    for(ri i=k;i>n;i--)ans=ans*i%p;
    ll tmp=1;
    for(ri i=k-n;i>=2;i--)tmp=tmp*i%p;
    ans=ans*fermat(tmp)%p;
    printf("%lld
",ans);
    return 0;
}