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;
}
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;
}
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~