preface

I found a big guy to go whoring to a number to play the game. Some questions were not written or I didn't write them.
Currently only: 10041005100710081011

Topic

Title Link: https://acm.hdu.edu.cn/contests/contest_show.php?cid=990

1004 Link with Balls

General idea of the topic

There are two kinds of boxes n n n, the color of the ball in each box is different

1. First, second i i i boxes can be taken out i k ( k ∈ N ) ik(k\in N) ik(k ∈ N) balls
2. Second kind i i i boxes can be taken out no more than i i i balls

Seek out m m Number of schemes for m balls.

1 ≤ T ≤ 1 0 5 , 1 ≤ n , m ≤ 1 0 6 1\leq T\leq 10^5,1\leq n,m\leq 10^6 1≤T≤105,1≤n,m≤106

Problem solving ideas

Derived from the generating function, the function of the first kind of ball is 1 − x i 1 − x \frac{1-x^i}{1-x} 1 − x1 − xi, the function of the second ball is 1 1 − x i \frac{1}{1-x^i} 1 − xi1, it is found that multiplication can offset. The final multiplication is
( 1 − x n ) ( 1 1 − x ) n + 1 (1-x^n)(\frac{1}{1-x})^{n+1} (1−xn)(1−x1​)n+1
And then come back
( n + m n ) − ( m − 1 n ) \binom{n+m}{n}-\binom{m-1}{n} (nn+m​)−(nm−1​)
Just fine

code

#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
const ll N=2e6+10,P=1e9+7;
ll T,inv[N],fac[N],n,m;
ll C(ll n,ll m)
{return fac[n]*inv[m]%P*inv[n-m]%P;}
signed main()
{
	inv[1]=1;
	for(ll i=2;i<N;i++)inv[i]=P-inv[P%i]*(P/i)%P;
	inv[0]=fac[0]=1;
	for(ll i=1;i<N;i++)fac[i]=fac[i-1]*i%P,inv[i]=inv[i-1]*inv[i]%P;
	scanf("%lld",&T);
	while(T--){
		scanf("%lld%lld",&n,&m);
		ll ans=C(m+n,n);
		m-=n+1;
		if(m>=0)ans=(ans-C(m+n,n)+P)%P;
		printf("%lld\n",ans);
	}
}

1005 Link with EQ

General idea of the topic

n n n Lattice stools. At the beginning, the first student will randomly choose a position to sit down, and the rest will choose a position farthest from the existing students to sit down. Ask for no students around the place. Look at how many students have sat down.
1 ≤ T ≤ 1 0 5 , 1 ≤ n ≤ 1 0 6 1\leq T\leq 10^5,1\leq n\leq 10^6 1≤T≤105,1≤n≤106

Problem solving ideas

set up f i f_i fi ， means only i i How many people can i do when i have a grid and there are students on each side? This can be very easy d p dp dp come out.

Then mainly consider the seat of the first person, judge the head and tail positions, and then the rest are right f f f find a prefix sum and you can calculate it quickly.

Time complexity O ( n + T log ⁡ P ) O(n+T\log P) O(n+TlogP)

code

#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
const ll N=1048576,P=1e9+7;
ll T,n,f[N],g[N],s[N];
ll power(ll x,ll b){
	ll ans=1;
	while(b){
		if(b&1)ans=ans*x%P;
		x=x*x%P;b>>=1;
	}
	return ans;
}
signed main()
{
	for(ll i=3;i<N;i++){
		ll mid=(i+1)/2;
		f[i]=f[mid-1]+f[i-mid]+1;
		s[i]=(s[i-1]+f[i]*2)%P;
	}
	for(ll i=1;i<N;i++)g[i]=f[i]+2;
	scanf("%lld",&T);
	while(T--){
		scanf("%lld",&n);
		if(n<=2){puts("1");continue;}
		if(n==3){puts("666666673");continue;}
		ll ans=(3*(n-4)+s[n-4])%P;
		(ans+=g[n-2]*2+g[n-3]*2)%=P;
		ans=ans*power(n,P-2)%P;
		printf("%lld\n",ans);
	}
	return 0;
}

1007 Link with Limit

General idea of the topic

Give permutation f f f. Then f i ( x ) f_i(x) fi (x) indicates x x x substitution i i Position after i times.
definition
g ( x ) = lim ⁡ n − > + ∞ 1 n ∑ i = 1 n f i ( x ) g(x)=\lim_{n->+\infty}\frac{1}{n}\sum_{i=1}^nf_i(x) g(x)=n−>+∞lim​n1​i=1∑n​fi​(x)

Ask whether for all g ( x ) ( x ∈ [ 1 , n ] ) g(x)(x\in [1,n]) g(x)(x ∈ [1,n]) are the same value.

1 ≤ n ≤ 1 0 5 1\leq n\leq 10^5 1≤n≤105

Problem solving ideas

Finally, it must be replaced into a ring. Consider whether the average value of each ring is equal

Time complexity O ( n ) O(n) O(n)

Code by our great stoorz \text{stoorz} stoorz wrote, so I didn't

1011 Yiwen with Formula

General idea of the topic

give n n A reproducible set of n numbers a a a. Find the product of the sum of all its subsets. model 998244353 998244353 998244353.

1 ≤ T ≤ 10 , 1 ≤ n ≤ 1 0 5 , ∑ n ≤ 2.5 × 1 0 5 , ∑ a i ≤ 4 × 1 0 5 1\leq T\leq 10,1\leq n\leq 10^5,\sum n\leq 2.5\times 10^5,\sum a_i\leq 4\times 10^5 1≤T≤10,1≤n≤105,∑n≤2.5×105,∑ai​≤4×105

Problem solving ideas

Divide and conquer violence N T T NTT NTT calculates the number of schemes for each sum, and then modulo because it is exponential φ ( 998244353 ) \varphi(998244353) φ (998244353) therefore, it is necessary to use any modulus to rush the grass.

code

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define ll long long
using namespace std;
const ll N=4e5*4+10,sqq=32768,p=998244352,P=998244353;
const double Pi=acos(-1);
struct complex{
    double x,y;
    complex (double xx=0,double yy=0)
    {x=xx;y=yy;return;}
}A[N],B[N],C[N],D[N];
struct Poly{
	ll a[N],n;
}F[20];
ll power(ll x,ll b,ll P){
	ll ans=1;
	while(b){
		if(b&1)ans=ans*x%P;
		x=x*x%P;b>>=1;
	}
	return ans;
}
complex operator+(complex a,complex b)
{return complex(a.x+b.x,a.y+b.y);}
complex operator-(complex a,complex b)
{return complex(a.x-b.x,a.y-b.y);}
complex operator*(complex a,complex b)
{return complex(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);}
complex w[N];
ll n,m,T,u[N],v[21],r[N];
void FFT(complex *f,ll op,ll n){
    for(ll i=0;i<n;i++)
        if(i<r[i])swap(f[i],f[r[i]]);
    for(ll p=2;p<=n;p<<=1){
        ll len=p>>1;
        for(ll k=0;k<n;k+=p)
            for(ll i=k;i<k+len;i++){
                complex tmp=w[n/len*(i-k)];
                if(op==-1)tmp.y=-tmp.y;
                complex tt=f[i+len]*tmp;
                f[i+len]=f[i]-tt;
                f[i]=f[i]+tt;
            }
    }
    if(op==-1){
        for(ll i=0;i<n;i++)
            f[i].x=(ll)(f[i].x/n+0.49);
    }
    return;
}
void MTT(ll *a,ll *b,ll *c,ll m,ll k){
    ll n=1;
    while(n<=m+k)n<<=1;
    for(ll i=0;i<n;i++){
        r[i]=(r[i>>1]>>1)|((i&1)?(n>>1):0);
    	A[i].x=A[i].y=B[i].x=B[i].y=0;
    	C[i].x=C[i].y=D[i].x=D[i].y=0;
	}
    for(ll len=1;len<n;len<<=1)
        for(ll i=0;i<len;i++)
            w[n/len*i]=complex(cos(i*Pi/len),sin(i*Pi/len));
    for(ll i=0;i<m;i++)A[i].x=a[i]/sqq,B[i].x=a[i]%sqq;
    for(ll i=0;i<k;i++)C[i].x=b[i]/sqq,D[i].x=b[i]%sqq;
    FFT(A,1,n);FFT(B,1,n);FFT(C,1,n);FFT(D,1,n);
    complex t1,t2;
    for(ll i=0;i<n;i++){
        t1=A[i]*C[i];t2=B[i]*D[i];
        B[i]=A[i]*D[i]+B[i]*C[i];
        A[i]=t1;C[i]=t2;
    }
    FFT(A,-1,n);FFT(B,-1,n);FFT(C,-1,n);
    for(ll i=0;i<n;i++){
    	c[i]=0;
        (c[i]+=(ll)(A[i].x)*sqq%p*sqq%p)%=p;
        (c[i]+=(ll)(B[i].x)*sqq%p)%=p;
        (c[i]+=(ll)(C[i].x))%=p;
    }
    return;
}
void Mul(Poly &a,Poly &b){
	MTT(a.a,b.a,a.a,a.n,b.n);
	a.n=a.n+b.n-1;return;
}
ll findq(){
	for(ll i=0;i<20;i++)
		if(!v[i]){v[i]=1;return i;}
}
ll Solve(ll l,ll r){
	if(l==r){
		ll p=findq();
		for(ll i=0;i<=u[l];i++)
			F[p].a[i]=0;
		F[p].a[0]=1;F[p].a[u[l]]=1;
		F[p].n=u[l]+1;return p;
	}
	ll mid=(l+r)>>1;
	ll ls=Solve(l,mid),rs=Solve(mid+1,r);
	Mul(F[ls],F[rs]);v[rs]=0;return ls;
}
signed main(){
	scanf("%lld",&T);
	while(T--){
		scanf("%lld",&n);
		bool flag=0;ll ans=1,sum=0;
		for(ll i=1;i<=n;i++){
			scanf("%lld",&u[i]);
			flag|=!u[i];sum+=u[i];
		}
		if(flag){puts("0");continue;}
		ll id=Solve(1,n);u[id]=0;
		for(ll i=1;i<=sum;i++)
			ans=ans*power(i,F[id].a[i],P)%P;
		printf("%lld\n",ans);
	}
}