2021牛客寒假算法基础集训营2，签到题FHIJ

F 牛牛与交换排序

//题目的限制条件是每次选翻转的区间只能更靠右,所以不能用长度为2的区间翻转两次达到把x翻到后面。所以确定k的方法是从左到右找到第一个顺序不对的数字，然后看它的位置和现在位置的距离，因为它一定是由一个这么大的区间翻过来的。
//接下来就按照给定的k去翻，直到不成立为止
#include<bits/stdc++.h>
using namespace std;
const int maxn = 100010;
int n, a[maxn], pos[maxn];
int main(){
	ios::sync_with_stdio(false);
	cin>>n;
	for(int i = 1; i <= n; i++){
		cin>>a[i];  pos[a[i]] = i;
	}
	int k = 0;
	for(int i = 1; i <= n; i++)
		if(pos[i]-i!=0){k=pos[i]-i; break;}
	for(int i = 1; i <= n; i++){
		if(a[i]==i)continue;
		if(a[i+k]!=i){
			cout<<"no"; return 0;
		}
		reverse(a+i,a+pos[i]+1);
	}
	cout<<"yes\n"<<k+1<<"\n";
	return 0;
}


  

 

  
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H 牛牛与棋盘

#include<bits/stdc++.h>
using namespace std;
int gcd(int a, int b){return !b?a:gcd(b,a%b);}
int main(){
	int n;  cin>>n;
	for(int i = 1; i <= n; i++){
		bool x;
		if(i%2==1)x = 0;
		else x = 1;
		for(int j = 1; j <= n; j++){
			cout<<x; x = !x;
		}
		cout<<"\n";
	}
	return 0;
}

  

 

  
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I 牛牛的“质因数”

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 4e6+10;
const LL mod = 1e9+7;
bitset<maxn>p;
LL pows[11] = {1,10,100,1000,10000,100000,1000000,10000000,100000000,1000000000,10000000000};
LL ma[maxn];
int main(){
	ios::sync_with_stdio(false);
	int n;  cin>>n;
	
	//map<int,string>ma;
	//map<LL,LL>ma;
	for(int i = 2; i <= n; i++){
		if(p[i])continue;
		//ma[i] = to_string(i);
		ma[i] = i;
		for(int j = 2*i; j <= n; j+=i){
			p[j] = 1;
			int k = 0, kk = i;
			while(kk > 0){k++; kk /= 10;}
			int t = j;
			while(t%i==0){
				//ma[j] += to_string(i);
				ma[j] = (ma[j]*pows[k]%mod+i)%mod;
				t /= i;
			}
		}
	}
	
	LL ans = 0;
	//for(auto i : ma){
	for(int i = 2; i <= n; i++){
		//cout<<i.first<<" "<<stoi(i.second)<<"\n";
		//ans += stoi(i.second)%mod;
		//cout<<i.first<<" "<<i.second<<"\n";
		//ans = (ans+i.second)%mod;
		ans = (ans+ma[i])%mod;
	}
	cout<<ans%mod<<"\n";
	return 0;
}


  

 

  
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J 牛牛想要成为hacker

//不能组成三角形不就好了??
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
int f[110];
int main(){
	int n;  cin>>n;
	f[0] = 1;  f[1] = 2;  cout<<"2 ";
	for(int i = 2; i <= min(40,n); i++){
		f[i] = f[i-1]+f[i-2];
		cout<<f[i]<<" ";
	}
	for(int i = 41; i <= n; i++)
		cout<<"1 ";
	return 0;
}


  

 

  
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文章来源: gwj1314.blog.csdn.net，作者：小哈里，版权归原作者所有，如需转载，请联系作者。

原文链接：gwj1314.blog.csdn.net/article/details/113663153