AtCoder Beginner Contest 158
地址:https://atcoder.jp/contests/abc158/tasks
A - Station and Bus
题意:AB之间通车,给三个字符,问能否通车,水题,只要不全是A 或者全是B都能通车
#include<bits/stdc++.h>
#include<vector>
#include<queue>
using namespace std;
typedef long long ll;
const int maxn = 3e5+10;
int main()
{
int a = 0;
int b = 0;
for(int i=0;i<3;i++)
{
char t;
cin>>t;
if(t=='A') a++;
else b++;
}
if(a==0||b==0) cout<<"No"<<endl;
else cout<<"Yes"<<endl;
return 0;
}
B - Count Balls
题意:循环
#include<bits/stdc++.h>
#include<vector>
#include<queue>
using namespace std;
typedef long long ll;
const int maxn = 3e5+10;
int main()
{
ll n,a,b;
cin>>n>>a>>b;
ll t = a+b;
ll time = n/t;
ll y = n%t;
ll sum = 0;
sum+=time*a;
if(y<=a) sum+=y;
else sum+=a;
cout<<sum<<endl;
return 0;
}
C - Tax Increase
题意:暴力枚举,1<=A<=B<=100 所有枚举到 1500(至少1250)
#include<bits/stdc++.h>
#include<vector>
#include<queue>
using namespace std;
typedef long long ll;
const int maxn = 3e5+10;
int main()
{
int a,b;
cin>>a>>b;
int flag = 0;
int ans = 0;
for(int i=1;i<=1500;i++)
{
if(floor(i*0.08)==a&&floor(i*0.1)==b)
{
ans = i;
flag = 1;
break;
}
}
if(flag==0) cout<<-1<<endl;
else cout<<ans<<endl;
return 0;
}
D - String Formation
题意:对字符串进行如下操作:反转,在首部插入字母,在尾部插入字母
思路:数据范围较大,直接模拟会超时,可以用list,它可以快速的插入和删除元素。
一直反转会超时:其实反转后向首插入元素,相当于反转前向尾部插入元素,按照这种方法就不需要反转了,但是要注意最后输出时,如果反转了偶数次,则正序输出,如果反转了奇数次则需要逆序输出。
#include<bits/stdc++.h>
#include<vector>
#include<queue>
#include<string>
#include<list>
using namespace std;
typedef long long ll;
const int maxn = 3e5+10;
int main()
{
list<char> q;
string s;
cin>>s;
int n;
cin>>n;
q.assign(s.begin(),s.end());
int flag = 0;
while(n--)
{
int t;
cin>>t;
if(t==1) flag++;
else
{
int f;
char c;
cin>>f>>c;
if(f==1)
{
if(flag%2==0)
{
q.push_front(c);
}
else
{
q.push_back(c);
}
}
else
{
if(flag%2!=0)
{
q.push_front(c);
}
else
{
q.push_back(c);
}
}
}
}
if(flag%2==0)
{
for(auto it = q.begin();it!=q.end();it++)
{
cout<<*it;
}
}
else
{
auto it = q.end();
for(it ;it!=q.begin();it--)
{
if(it==q.end()) continue;
cout<<*it;
}
cout<<*it;
}
cout<<endl;
return 0;
}