A - Spread
输入字符串,字符之间加上空格输出
B - Next
输出数组当中第二大的数
C - Count xxx
统计每个字符出现过的最长长度,再累加即可
cpp
#include<bits/stdc++.h>
#pragma GCC optimize("Ofast")
#define INF 0x3f3f3f3f
#define IOS ios::sync_with_stdio(false);cin.tie(0);
#define int long long
#define pb push_back
#define vct vector
#define checkbit __builtin_popcount
#define gcd __gcd
#define use int T;cin>>T;while(T--)
#define LEN length()
#define all(a) a.begin(),a.end()
template<class T> bool mmax(T &u, T v) { return u < v ? (u = v, 1) : 0; }
template<class T> bool mmin(T &u, T v) { return u > v ? (u = v, 1) : 0; }
#define lowbit(x) (x&(-x))
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endl
using namespace std;
typedef pair<int,int>pii;
signed main()
{IOS
int n;cin>>n;
string a;
cin>>a;
vct<int>cnt(35,0);
vct<int>mas(35,0);
if(cnt[a[0]-'a']==0)
{
cnt[a[0]-'a']++;
mmax(mas[a[0]-'a'],cnt[a[0]-'a']);
}
for(int i=1;i<n;i++){
if(cnt[a[i]-'a']==0)
{
cnt[a[i]-'a']++;
mmax(mas[a[i]-'a'],cnt[a[i]-'a']);
}
if(a[i]==a[i-1]){
cnt[a[i]-'a']++;
}
else {
mmax(mas[a[i-1]-'a'],cnt[a[i-1]-'a']);
cnt[a[i-1]-'a']=0;
}
}int ans=0;
mmax(mas[a[n-1]-'a'],cnt[a[n-1]-'a']);
for(int i=0;i<35;i++){
ans+=mas[i];
}
cout<<ans<<endl;
return 0;
}
D - Election Quick Report
给定数组,每次第次的票加一,每次都输出最多票的人,
我们用记录当前最大票数的人,在第次投票时,答案只有和两个人,每次视情况输出,并且更新的值即可.
cpp
#include<bits/stdc++.h>
#pragma GCC optimize("Ofast")
#define INF 0x3f3f3f3f
#define IOS ios::sync_with_stdio(false);cin.tie(0);
#define int long long
#define pb push_back
#define vct vector
#define checkbit __builtin_popcount
#define gcd __gcd
#define use int T;cin>>T;while(T--)
#define LEN length()
#define all(a) a.begin(),a.end()
template<class T> bool mmax(T &u, T v) { return u < v ? (u = v, 1) : 0; }
template<class T> bool mmin(T &u, T v) { return u > v ? (u = v, 1) : 0; }
#define lowbit(x) (x&(-x))
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endl
using namespace std;
typedef pair<int,int>pii;
bool cmp(int a,int b){
return a>b;
}
signed main()
{IOS
int n,m;cin>>n>>m;
vct<int>a(m+1);
vct<int>cnt(n+1);cin>>a[1];
int mas=a[1];cout<<a[1]<<endl;
cnt[a[1]]++;
for(int i=2;i<=m;i++){
cin>>a[i];
cnt[a[i]]++;
if(a[i]!=mas){
if(cnt[a[i]]>cnt[mas]){
mas=a[i];
printf("%lld\n",mas);
}
else if(cnt[a[i]]==cnt[mas]&&a[i]<mas){
mas=a[i];
printf("%lld\n",mas);
}
else {
printf("%lld\n",mas);
}
}else{
printf("%lld\n",mas);
}
}
return 0;
}
E - Stamp
给一张空白的纸,一个印章,问是否可以印成的样子(印章每次会覆盖重复的部分)
利用BFS搜索
cpp
#include<bits/stdc++.h>
#pragma GCC optimize("Ofast")
#define INF 0x3f3f3f3f
#define IOS ios::sync_with_stdio(false);cin.tie(0);
#define int long long
#define pb push_back
#define vct vector
#define checkbit __builtin_popcount
#define gcd __gcd
#define use int T;cin>>T;while(T--)
#define LEN length()
#define all(a) a.begin(),a.end()
template<class T> bool mmax(T &u, T v) { return u < v ? (u = v, 1) : 0; }
template<class T> bool mmin(T &u, T v) { return u > v ? (u = v, 1) : 0; }
#define lowbit(x) (x&(-x))
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endl
using namespace std;
typedef pair<int,int>pii;
const int N= 2e5+7;
signed main()
{IOS
int n,m;cin>>n>>m;
string s,t;cin>>s>>t;
vct<bool>st(N);
queue<int> q;
for(int i=0;i+m-1<n;i++){
bool flag=1;
for(int j=0;j<m;j++){
if(s[i+j]!=t[j])flag=0;
}
if(flag)q.push(i),st[i]=1;
}
while(!q.empty()){
int u=q.front();q.pop();
for(int j=0;j<m;j++)s[u+j]='#';
for(int i=max(u-m+1,0*1ll);i<=u+m-1&&i+m-1<n;i++){
if(!st[i])
{int flag=1;
for(int j=0;j<m;j++){
if(s[i+j]!='#'&&s[i+j]!=t[j])flag=0;
}
if(flag)q.push(i),st[i]=1;
}
}
}
bool isok=1;
for(int i=0;i<n;i++){
if(s[i]!='#')isok=0;
}
if(isok)cout<<"Yes"<<endl;
else cout<<"No"<<endl;
return 0;
}
F - Colored Ball
每次操作将a位置当中的元素,放到b位置,因为是颜色,故用平衡树
如果按照题目的要求直接写不加优化的话亲测TLE
故当a当中的元素多于b的时候,交换a,b当中的元素,再插入a,效果相同,这样可以达到最优效果
cpp
#include<bits/stdc++.h>
#pragma GCC optimize("Ofast")
#define INF 0x3f3f3f3f
#define IOS ios::sync_with_stdio(false);cin.tie(0);
#define int long long
#define pb push_back
#define vct vector
#define checkbit __builtin_popcount
#define gcd __gcd
#define use int T;cin>>T;while(T--)
#define LEN length()
#define all(a) a.begin(),a.end()
template<class T> bool mmax(T &u, T v) { return u < v ? (u = v, 1) : 0; }
template<class T> bool mmin(T &u, T v) { return u > v ? (u = v, 1) : 0; }
#define lowbit(x) (x&(-x))
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endl
using namespace std;
typedef pair<int,int>pii;
const int N= 2e5+7;
signed main()
{IOS
int n,q;cin>>n>>q;
int x;
set<int>tot[n+1];
for(int i=1;i<=n;i++)
{
cin>>x;
tot[i].insert(x);
}
while(q--){
int a,b;cin>>a>>b;
if(tot[a].size()>tot[b].size())
swap(tot[a],tot[b]);
for(auto z:tot[a]){
tot[b].insert(z);
}
tot[a].clear();
cout<<tot[b].size()<<"\n";
}
return 0;
}