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哦呼!20211127 小测

作者:互联网

A bzoj - 3260 跳

简单且狗儿题,最多算根号次,不取模差不多得了。

//how to get this one????
#include<bits/stdc++.h>
const int mod=1e9+7;
long long n,m;
long long qpow(long long bas,int times)
{
	long long res=1;
	for(; times; times>>=1,bas=bas*bas%mod)
	{
		if(times&1)	res=res*bas%mod;
	}
	return res;
}
long long com(long long one,long long ano)
{
	long long resx=1,resy=1;
	for(int i=0; i<ano; ++i)
	{
		resx=resx*((one-i+mod)%mod)%mod;
		resy=resy*((i+1)%mod)%mod;
	}
	return resx*qpow(resy,mod-2)%mod;
}
signed main()
{
	freopen("jump.in","r",stdin);
	freopen("jump.out","w",stdout);
	scanf("%lld%lld",&n,&m);
	if(n>m)	n^=m^=n^=m;
	printf("%lld\n",(com(n+m+1,n)+m)%mod);
	fprintf(stderr,"%lld %lld",n,m);
	return 0;
}

B haoi - 2010 计数

其实就是康托展开,把字符串中的 \(\texttt{0}\) 全部放到开头算全排列,然后答案即 \(\displaystyle\prod_{i=0}^{9}\binom{m-\sum_{0\leqslant j<i}c_j}{c_i}\),\(c\) 是桶。

#include<bits/stdc++.h>
using namespace std;
int val[100],cnt[20],n;
long long arrcom[2100][2100],ans;
long long com(int one,int ano)
{
	if(arrcom[one][ano])	return arrcom[one][ano];
	else if(one<ano)	return arrcom[one][ano]=0;
	else if(ano==0 || one==ano)	return 1;
	return arrcom[one][ano]=com(one-1,ano-1)+com(one-1,ano);
}
signed main()
{
	freopen("count.in","r",stdin);
	freopen("count.out","w",stdout);
	string fuck;
	cin>>fuck;
	for(int i=0; i<int(fuck.size()); ++i)	val[++n]=fuck[i]-'0';
	for(int i=1; i<=n; ++i)	cnt[val[i]]++;
	for(int i=1; i<=n; ++i)
	{
		for(int j=0; j<val[i]; ++j)
		{
			if(cnt[j])
			{
				cnt[j]--;
				long long tmp=1;
				int up=n-i;
				for(int k=0; k<10; ++k)
				{
					if(cnt[k])
					{
						tmp*=com(up,cnt[k]);
						up-=cnt[k];
					}
				}
				ans+=tmp;
				cnt[j]++;
			}
		}
		cnt[val[i]]--;
	}
	printf("%lld\n",ans);
	return 0;
}

C acmhdu - 6397 Character Encoding

我也不知道我为什么就是推错了,可能是我没长脑子吧。你看着这个题是不是很想求 \(f(z)\) 表示至少 \(z\) 个超限的方案数,然后答案就是 \(\displaystyle\sum_{i=0}^{\lfloor\frac{k}{n}\rfloor}(-1)^if(i)\)。\(f(i)\) 表示出来就是 \(\binom{m}{i}\binom{k-n\times i+m-1}{m-1}\)。

//i'm the shabbiest one ever
#include<bits/stdc++.h>
template <int P>
struct Z {
  int x;
  Z(const int a = 0) : x(norm(a)) {}
  static int norm(const int& t) {
    if (t < 0) return t + P;
    if (t >= P) return t - P;
    return t;
  }
  Z inv() const { return assert(x), power(x, P - 2); }
  static Z power(Z x, long long y) {
    Z res = 1;
    for (; y; y >>= 1, x *= x)
      if (y & 1) res *= x;
    return res;
  }
  int val() const { return x; }
  Z operator-() { return norm(-x); }
  friend Z operator+(const Z& a, const Z& b) { return Z(norm(a.val() + b.val())); }
  friend Z operator-(const Z& a, const Z& b) { return Z(norm(a.val() - b.val())); }
  friend Z operator*(const Z& a, const Z& b) { return Z(static_cast<long long>(a.val()) * b.val() % P); }
  friend Z operator/(const Z& a, const Z& b) { return a * b.inv(); }
  Z &operator+=(const Z& t) { return (*this) = (*this) + t; }
  Z &operator-=(const Z& t) { return (*this) = (*this) - t; }
  Z &operator*=(const Z& t) { return (*this) = (*this) * t; }
  Z &operator/=(const Z& t) { return (*this) = (*this) / t; }
  static int mod() { return P; }
};
using mint=Z<998244353 >;
struct Simple {
  std::vector<mint> fac, ifac;
  Simple() : fac(1, 1), ifac(1, 1) {}
  mint gfac(int n) { return check(n), fac[n]; }
  mint gifac(int n) { return check(n), ifac[n]; }
  void check(int n) {
    int pn = fac.size();
    for (int i = pn; i <= n; ++i) fac.emplace_back(fac.back() * i);
    for (int i = pn; i <= n; ++i) ifac.emplace_back(fac[i].inv());
  }
  mint binom(int n, int k) {
    assert(n >= k), check(n);
    return fac[n] * ifac[n - k] * ifac[k];
  }
} simp;
mint C(int x,int y){return simp.binom(x,y);
}
signed main()
{
	freopen("encoding.in","r",stdin);
	freopen("encoding.out","w",stdout);
	int T,n,m,k;
	scanf("%d",&T);
	for(; T--;)
	{
		scanf("%d%d%d",&n,&m,&k);
		if(k/m>n-1)	{puts("0");continue;}
		mint ans=0;
        for(int i=0;i<=k/n;i++)
        {
            if(i&1)ans=(ans-(C(m,i)*C(k-n*i+m-1,m-1)));
            else ans=(ans+(C(m,i)*C(k-n*i+m-1,m-1)));
        }
		printf("%d\n",ans.val());
	}
	return 0;
}

D Non-Resource

萌萌题,我打 std::vector 螺旋升天。

就嗯上容斥,std::map 维护即可。

#include<bits/stdc++.h>
using namespace std;
struct node
{
	int a[5];
	void get(int x,int y,int z,int w,int h)
	{
		int cur=0;
		for(int i:{x,y,z,w,h})	a[cur++]=i;
	}
	int& operator[](const int i)
	{
		return a[i];
	}
	friend bool operator<(node one,node ano)
	{
		for(int i=0; i<5; ++i)
		{
			if(one[i]<ano[i])	return true;
			else if(one[i]>ano[i])	return false;
		}
		return false;
	}
}emp;
map<node,int> mp[6];
int n;
long long ans;
signed main()
{
	freopen("against.in","r",stdin);
	freopen("against.out","w",stdout);
	scanf("%d",&n);
	for(int i=1,b[5]; i<=n; ++i)
	{
		for(int i=0; i<5; ++i)	scanf("%d",&b[i]);
		sort(b,b+5);
		for(int S=1; S<(1<<5); ++S)
		{
			int tot=0;
			node tmp=emp;
			for(int j=0; j<5; ++j)
			{
				if(S&(1<<j))	tmp[tot++]=b[j];
			}
			mp[tot][tmp]++;
		}
	}
	ans=1ll*n*n;
	for(int i=1; i<6; ++i)
	{
		for(auto it=mp[i].begin(); it!=mp[i].end(); ++it)	ans+=((i&1)?-1ll:1ll)*(it->second)*(it->second);
	}
	printf("%lld\n",ans/2);
	return 0;
}

标签:return,val,int,long,operator,小测,20211127,const
来源: https://www.cnblogs.com/orchid-any/p/15611818.html