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AcWing 886. 费马小定理, 快速幂, 求组合数 II  

思路讲解

https://www.acwing.com/solution/content/16482/

这个题解非常好,我的快速幂以及逆元都是参考他的题解。

AcWing 876. 快速幂求逆元

AcWing 875. 快速幂

总的来说思路是很简单的,就是预处理公式法。但是除法这取模就会出问题,所以需要用逆元。

AC代码

https://www.acwing.com/problem/content/submission/code_detail/40914515/

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// Problem: 求组合数 II
// Contest: AcWing
// URL: https://www.acwing.com/problem/content/888/
// Memory Limit: 64 MB
// Time Limit: 1000 ms
// by znzryb
//
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define all(x) x.begin(),x.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define fi first
#define se second
#define pb push_back
#define SZ(a) ((int) a.size())

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
typedef pair<DB,DB> pdd;
typedef pair<ll,bool> plb;
constexpr ll MAXN=static_cast<ll>(1e5)+10,INF=static_cast<ll>(5e18)+3;
constexpr ll mod=static_cast<ll>(1e9)+7;

ll N,M,T;
ll fact[MAXN],infact[MAXN];

ll binpow(ll a,ll k,const ll &p){ // 迭代快速幂
ll res=1;
while(k>0){
if((k&1)==1) res=a*res%p;
a=a*a%p;
k>>=1;
}
return res;
}

inline void solve(){
ll a,b;
cin>>a>>b;
cout<<(fact[a]*infact[b]%mod*infact[a-b])%mod<<"\n";
}

int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
fact[0]=1;
infact[0]=binpow(fact[0],mod-2,mod);;
FOR(i,1,MAXN-5){
fact[i]=(fact[i-1]*i)%mod;
infact[i]=binpow(fact[i],mod-2,mod);
}
#ifdef LOCAL
FOR(i,1,15){
cerr<<fact[i]<<" ";
}
cerr<<"\n";
FOR(i,1,15){
cerr<<infact[i]<<' ';
}
cerr<<"\n";
#endif
cin>>T;
while(T--){
solve();
}
// solve();
return 0;
}
/*
AC
https://www.acwing.com/problem/content/submission/code_detail/40914515/
*/

心路历程(WA,TLE,MLE……)