Codeforces Round #606 (Div. 2, based on Technocup 2020 Elimination Round 4)

链接

签到题，求出位数，然后9*（位数-1）+ 从位数相同的全一开始加看能加几次的个数

#include<bits/stdc++.h>

 using namespace std;

 int main(int argc, char const *argv[])

 {

 	int t;

 	int y;

 	cin>>t;

 	int ans = 0;

 	while(t--)

 	{

 		cin>>y;

 		ans = 0;

 		int weishu = 0;

 		int temp = y;

 		while(temp>0)

 		{

 			weishu++;

 			temp/=10;

 		}

 		ans += (weishu-1)*9;

 		temp = 1;

 		for (int i = 1; i < weishu; ++i)

 		{

 			temp = temp*10+1;

 		}

 		for (int i = temp; i <= y; i += temp)

 		{

 			ans++;

 		}

 		cout<<ans<<endl;

 	}

 	return 0;

 }

链接

思路，用堆来维护所有的偶数，每次取最大的来除以二，但是需要在处理的时候去重。用STL的优先队列就可

#include<bits/stdc++.h>

using namespace std;

int main(int argc, char const *argv[])

{

	//数的数量啊

	//排序一波？

	priority_queue<int> q;

	int t;

	cin>>t;

	while(t--)

	{

		int n;

		cin>>n;

		int num;

		int flag = 0;

		for (int i = 0; i < n; ++i)

		{

			cin>>num;

			if (!(num%2)) //不是奇数，插入就行

			{

				q.push(num);

				flag++;

			}

		}

		if(flag)

		{

			int top = q.top();

			int ans = 0;

			int cur = 0;

			q.pop();

			ans++;

			if (!((top/2)%2))

			{

				q.push(top/2);

			}

			while(!q.empty())

			{

				cur = q.top();

				//cout<<cur<<endl;

				q.pop();

				if (cur == top)

					continue;

				else  //不是重复了哈

				{

					top = cur;

					ans++;

					if (!((top/2)%2))

					{

						q.push(top/2);

					}

				}

			}

			cout<<ans<<endl;

		}

		else

			cout<<"0"<<endl;

	}

	return 0;

}

链接

思考一下，因为two和one,是可能出现连起来出现的，所以先扫描一遍所有的twone这种的，去掉中间的o,然后再扫描一次把剩下的one和two

去掉中间那个字母就行。

#include<bits/stdc++.h>

using namespace std;

char s[150001];

int del[150001];

 int main(int argc, char const *argv[])

 {

 	ios::sync_with_stdio(false);

 	std::vector<int> a;

 	int t;

 	cin>>t;

 	while(t--)

 	{

 		cin>>s;

 		int ans = 0;

 		a.clear();

 		//memset(del, 0, sizeof(del));

 		int len = strlen(s);

 		for (int i = 0; i < len; ++i)

 		{

 			del[i]=0;

 		}

 		for (int i = 2; i < len-2; ++i)

 		{

 			if (s[i]=='o')

 			{

 					if (s[i-2]=='t'&&s[i-1]=='w'&&s[i+1]=='n'&&s[i+2]=='e')

 					{

 						//s.erase(i,1);

 						a.push_back(i+1);

 						del[i] = 1;

 						//cout<<i<<" ";

 						++ans ;

 					}

 			}

 		}

 		for (int i = 1; i <=len-2; ++i)

 		{

 			if (s[i]=='w')

 				{

 					if (s[i-1]=='t'&&s[i+1]=='o'&&!del[i+1])

 					{

 						del[i]=1;

 						a.push_back(i+1);

 						++ans;

 						//cout<<i<<" ";

 					}

 				}

 				if (s[i]=='n')

 				{

 					if (s[i-1]=='o'&&s[i+1]=='e'&&!del[i-1])

 					{

 						del[i]=1;

 						a.push_back(i+1);

 						++ans;

 						//cout<<i<<" ";

 					}

 				}

 		}

 		cout<<ans<<endl;

 		for (int i = 0; i < ans; ++i)

 		{

 			cout<<a[i]<<' ';

 		}

 		cout<<endl;

 	}

 	return 0;

 }

未完待续~

巴特西

Codeforces Round #606 (Div. 2, based on Technocup 2020 Elimination Round 4)

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