杨辉三角（杨辉三角性质 + 组合数+求导数）

傷城~ 2022-09-10 10:16 135阅读 0赞

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上面的公式y取 1  
然后进行一堆构造，对比.

当mod 很多的时候，可以写一个mod 函数，（师兄说的）

mod和乘法的优先级，从左往右依次执行。

// Problem: 杨辉三角
    // Contest: NowCoder
    // URL: https://ac.nowcoder.com/acm/contest/20263/C
    // Memory Limit: 524288 MB
    // Time Limit: 2000 ms
    // 
    // Powered by CP Editor (https://cpeditor.org)
    
    #include<iostream>
    #include<cstdio>
    #include<string>
    #include<ctime>
    #include<cmath>
    #include<cstring>
    #include<algorithm>
    #include<stack>
    #include<climits>
    #include<queue>
    #include<map>
    #include<set>
    #include<sstream>
    #include<cassert>
    #include<bitset>
    #include<list>
    #include<unordered_map>
    using namespace std;
    #define ff first
    #define ss second
    #define lowbit(x) (x&-x)
    #define pf(a) printf("%d\n",a)
    #define mem(x,y) memset(x,y,sizeof(x))
    #define dbg(x) cout << #x << " = " << x << endl
    #define rep(i,l,r) for(int i = l; i <= r; i++)
    #define fep(i,a,b) for(int i=b; i>=a; --i)
    typedef pair<int,int> PII;
    typedef long long ll;
    typedef unsigned long long ull;
    
    
    const ll mod = 99824353;
    
    ll mul(ll a, ll b)
    { 
    	return ((a%mod) * (b%mod)) % mod;
    }
    
    ll qmi(ll a, ll b)
    { 
    	ll res = 1ll;
    	while(b)
    	{ 
    		if(b&1ll) res = mul(res, a);
    		a = mul(a, a);
    		b >>= 1ll;
    	}
    	return res;
    }
    
    
    void solve()
    { 
    	ll n;
    	cin >> n;
    	if(n==1) cout << 0 << endl;
    	else if(n==2) cout << 1 << endl;
    	else cout << mul (mul(n,n-1), qmi(2ll, n-3));
    }
    
    int main()
    { 
    	solve();
    	return 0;
    }

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