Codeforces Round # 631 (Div. 2) (D. XOR property conclusion E simulation?)

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Reference from: this

D. Dreamoon Likes Sequences

Topic:

Given a limit d and modulus mod, to find the number of arrays a that can be constructed to satisfy the condition, the following conditions need to be met:

        The length of array a is greater than or equal to 1 
        array a
        . The minimum value of strictly increasing array a is greater than or equal to 1, and the maximum value of array a is less than or equal to d.
        For array a, construct an array b: when
        i == 1: b [1] = a When [1]
        i> 1: b [i] = b [i-1] ^ a [i] 
        array b strictly increments

Practice: I am confused about the math idiot, I do n’t know where to start, so Baidu solves the problem

It's not easy to get this conclusion. This kind of mathematical foundation should be able to see this conclusion at a glance. It's still my dish

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=(b);++i)
#define mem(a,x) memset(a,x,sizeof(a))
#define pb push_back
#define pi pair<int, int>
#define mk make_pair
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
ll n,mod,ans,f[35];
int main()
{
    f[0]=1;
    for(int i=1;i<=30;++i) f[i]=f[i-1]*2;
    int _;cin>>_;while(_--)
    {
        scanf("%lld%lld",&n,&mod);
        ans=1;
        for(int i=0;i<=30;++i){
            if(f[i]>n) break;
            ans*=(min(f[i+1]-1,n)-(f[i]-1)+1);
            ans%=mod;
            ans+=mod;
            ans%=mod;
        }
        ans-=1;
        ans=(ans+mod)%mod;
        printf("%lld\n",ans);
    }
}

E. Drazil Likes Heap

I can't write it, the main question is too circumvent, search for the solution, and found the solution of this brother: this

what? This is called greed? I feel that this is called simulating according to the meaning of the question after understanding the meaning of the question

#include<bits/stdc++.h>
using namespace std;

typedef long long LL;

typedef unsigned long long ull;

const int inf=0x3f3f3f3f;

const int N=(1<<21)+100;

int a[N];

int get_id(int k)
{
	if(!a[k<<1]&&!a[k<<1|1])
		return k;
	if(a[k<<1]>a[k<<1|1])
		return get_id(k<<1);
	else
		return get_id(k<<1|1);
}

void dfs(int k)
{
	if(!a[k<<1]&&!a[k<<1|1])
		a[k]=0;
	else if(a[k<<1]>a[k<<1|1])
	{
		a[k]=a[k<<1];
		dfs(k<<1);
	}
	else
	{
		a[k]=a[k<<1|1];
		dfs(k<<1|1);
	}
}

int main()
{

	int w;
	cin>>w;
	while(w--)
	{
		int h,g;
		scanf("%d%d",&h,&g);
		for(int i=1;i<1<<(h+1);i++)
			a[i]=0;
		for(int i=1;i<1<<h;i++)
			scanf("%d",a+i);
		vector<int>ans;
		int limit=(1<<g)-1;
		for(int i=1;i<1<<g;i++)
			while(get_id(i)>limit)
			{
				ans.push_back(i);
				dfs(i);
			}
		LL sum=0;
		for(int i=1;i<1<<g;i++)
			sum+=a[i];
		printf("%lld\n",sum);
		for(int i=0;i<ans.size();i++)
			printf("%d ",ans[i]);
		puts("");
	}


    return 0;
}