A. ABC Swap
题意:
题解:
AC代码
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define endl '\n'
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
//const int mod=1e9+7;
const int mod=998244353;
const double eps = 1e-10;
const double pi=acos(-1.0);
const int maxn=1e6+10;
const ll inf=0x3f3f3f3f;
const int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int a,b,c;
cin>>a>>b>>c;
swap(a,b);
swap(a,c);
cout<<a<<' '<<b<<' '<<c<<endl;
return 0;
}
B.Popular Vote
题意:
题解:
AC代码
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define endl '\n'
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
//const int mod=1e9+7;
const int mod=998244353;
const double eps = 1e-10;
const double pi=acos(-1.0);
const int maxn=1e6+10;
const ll inf=0x3f3f3f3f;
const int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
int a[110];
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int n,m;
cin>>n>>m;
int sum=0;
for(int i=1;i<=n;i++){
cin>>a[i];
sum+=a[i];
}
sum=(sum+4*m-1)/(4*m);
int ans=0;
for(int i=1;i<=n;i++){
if(a[i]>=sum)ans++;
}
if(ans>=m)cout<<"Yes";
else cout<<"No";
return 0;
}
C.Replacing Integer
题意:
题解:
AC代码
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define endl '\n'
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
//const int mod=1e9+7;
const int mod=998244353;
const double eps = 1e-10;
const double pi=acos(-1.0);
const int maxn=1e6+10;
const ll inf=0x3f3f3f3f;
const int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
int a[110];
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
ll n,k;
cin>>n>>k;
n%=k;
cout<<min(n,k-n);
return 0;
}
D.Lunlun Number
题意:
题解:
AC代码
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define endl '\n'
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
//const int mod=1e9+7;
const int mod=998244353;
const double eps = 1e-10;
const double pi=acos(-1.0);
const int maxn=1e6+10;
const ll inf=0x3f3f3f3f;
const int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
queue<ll> q;
int n;
cin>>n;
int num=0;
ll ans;
for(int i=1;i<=9;i++) q.push(i);
while(!q.empty()){
ll p=q.front();
q.pop();
num++;
if(num==n){ans=p;break;}
int f=p%10;
if(f==0){
q.push(p*10);
q.push(p*10+1);
}
else if(f==9){
q.push(p*10+8);
q.push(p*10+9);
}
else{
q.push(p*10+f-1);
q.push(p*10+f);
q.push(p*10+f+1);
}
}
cout<<ans;
return 0;
}
E.Yutori
题意:
题解:
AC代码
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define endl '\n'
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
//const int mod=1e9+7;
const int mod=998244353;
const double eps = 1e-10;
const double pi=acos(-1.0);
const int maxn=2e5+10;
const ll inf=0x3f3f3f3f;
const int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
int l[maxn],r[maxn];
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int n,k,c;
cin>>n>>k>>c;
string s;
cin>>s;
int cnt=k-1,mx=n;
for(int i=s.length()-1;i>=0;i--)
if(s[i]=='o'&&i<mx&&cnt>=0)r[cnt]=i,mx=i-c,cnt--;
cnt=0,mx=-1;
for(int i=0;i<s.length();i++)
if(s[i]=='o'&&i>mx)l[cnt]=i,mx=i+c,cnt++;
for(int i=0;i<k;i++)
if(l[i]==r[i])cout<<l[i]+1<<endl;
return 0;
}
F.Division or Substraction
题意:
题解:
AC代码
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define endl '\n'
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
//const int mod=1e9+7;
const int mod=998244353;
const double eps = 1e-10;
const double pi=acos(-1.0);
const int maxn=1e6+10;
const ll inf=0x3f3f3f3f;
const int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
ll n,t;
cin>>n;
int ans=0;
for(ll i=1;i*i<=n-1;i++){
if((n-1)%i==0){
if(i!=1)ans++;
if((n-1)/i!=i)ans++;
}
}
for(ll i=2;i*i<=n;i++){
if(n%i==0){
t=n;
while(t%i==0)t/=i;
if(t%i==1)ans++;
}
}
cout<<ans+1;
return 0;
}