Good Bye 2021: 2022 is NEAR

A. Integer Diversity 题目：

思路分析：

就是给你一个序列通过改变数字的正负可以得到最大不同数字的个数是多少

代码分析：

#include 
using namespace std;
typedef long long ll;
typedef unsigned long long  ull;
inline int read() {
    int x=0,f=1; char c=getchar();
    while(c<'0'||c>'9') {if(c=='-') f=-1;c=getchar();}
    while(c>='0'&&c<='9') {x=(x<<1)+(x<<3)+c-'0';c=getchar();}
    return x*f;
}
ll _gcd(ll m, ll n){return n == 0 ? m : _gcd(n, m%n);}
ll _lcm(ll m, ll n){return m*n / _gcd(m, n);}
ll pows(ll base, ll power,ll mod){ll result=1;while(power>0){if(power&1){result=result*base%mod;}power>>=1;base=(base*base)%mod;}return result;}
ll poww(ll base, ll power){ll result=1;while(power>0){if(power&1){result=result*base;}power>>=1;base=(base*base);}return result;}
void init(){ios::sync_with_stdio(0);cin.tie(0),cout.tie(0);}
const int MAX=1e6+10;
int t;
bool cmp(int a,int b){return a>b;}
ll max(ll a,ll b){if(a>b)return a;else return b;}


void solve(){
    int n;cin>>n;
    int a[110];
    for(int i=0;i<=100;i++)a[i]=0;
    for(int i=0;i>x;a[abs(x)]++;}int ans=0;
    if(a[0]!=0) ans++;
    for(int i=1;i<=100;i++) {
        if(a[i]>=2) ans+=2;
        else if(a[i]==1) ans++;
    }
    cout<>t;
    while(t--) solve();
//    solve();
    return 0;
}

B. Mirror in the String 题目：

思路分析：

就是从字符串的第一位找一小段字符串然后镜像对称使其字典序最小

我们可以开头遍历如果遇到字符序不减则停止然后镜像对称即可

代码实现：

#include 
using namespace std;
typedef long long ll;
typedef unsigned long long  ull;
inline int read() {
    int x=0,f=1; char c=getchar();
    while(c<'0'||c>'9') {if(c=='-') f=-1;c=getchar();}
    while(c>='0'&&c<='9') {x=(x<<1)+(x<<3)+c-'0';c=getchar();}
    return x*f;
}
ll _gcd(ll m, ll n){return n == 0 ? m : _gcd(n, m%n);}
ll _lcm(ll m, ll n){return m*n / _gcd(m, n);}
ll pows(ll base, ll power,ll mod){ll result=1;while(power>0){if(power&1){result=result*base%mod;}power>>=1;base=(base*base)%mod;}return result;}
ll poww(ll base, ll power){ll result=1;while(power>0){if(power&1){result=result*base;}power>>=1;base=(base*base);}return result;}
void init(){ios::sync_with_stdio(0);cin.tie(0),cout.tie(0);}
const int MAX=1e6+10;
int t;
bool cmp(int a,int b){return a>b;}
ll max(ll a,ll b){if(a>b)return a;else return b;}


void solve(){
    int n;cin>>n;
    string s;s="";cin>>s;s="z"+s;
    int ans=1;
    if(s[1]==s[2]) ans=1;
    else {for(int i=1;i<=n;i++){
        if((s[i]-'a')>(s[i-1]-'a')){
            ans=i-1;break;
        }
        else ans=i;
        
    }}
//    cout<=1;i--){
        cout<>t;
    while(t--) solve();
//    solve();
    return 0;
}

C. Representative Edges 题目：

思路分析：

题意就是给你一个数字序列然后可以修改单个数字的值来构造一个等差序列

找出最少的操作数

由于数据很小我们可以枚举d 然后遍历得出最小的操作数就行

代码实现：

#include 
using namespace std;
typedef long long ll;
typedef unsigned long long  ull;
inline int read() {
    int x=0,f=1; char c=getchar();
    while(c<'0'||c>'9') {if(c=='-') f=-1;c=getchar();}
    while(c>='0'&&c<='9') {x=(x<<1)+(x<<3)+c-'0';c=getchar();}
    return x*f;
}
ll _gcd(ll m, ll n){return n == 0 ? m : _gcd(n, m%n);}
ll _lcm(ll m, ll n){return m*n / _gcd(m, n);}
ll pows(ll base, ll power,ll mod){ll result=1;while(power>0){if(power&1){result=result*base%mod;}power>>=1;base=(base*base)%mod;}return result;}
ll poww(ll base, ll power){ll result=1;while(power>0){if(power&1){result=result*base;}power>>=1;base=(base*base);}return result;}
void init(){ios::sync_with_stdio(0);cin.tie(0),cout.tie(0);}
const int MAX=1e6+10;
int t;
bool cmp(int a,int b){return a>b;}
ll max(ll a,ll b){if(a>b)return a;else return b;}


void solve(){
    int n;cin>>n;
    int a[110];
    for(int i=0;i>x;a[i]=x;}
    int ans=0x3f3f3f3f;
    if(n<=2) {cout<<0<t;
    while(t--) solve();
//    solve();
    return 0;
}

D. Keep the Average High 题目：

思路分析：

题意就是从给出的数学序列里找出x个数字然后这个x个数字的序列中任意区间和都满足一个关系区间元素不少于二个

我们可以1-n进行枚举当前第i个数能否选中和前面的数有关系

dp[i][0/1][0/1] 表示第i位前一位和第i位的选择情况

状态转移方程看代码吧！

代码实现：

#include 
using namespace std;
typedef long long ll;
typedef unsigned long long  ull;
inline int read() {
    int x=0,f=1; char c=getchar();
    while(c<'0'||c>'9') {if(c=='-') f=-1;c=getchar();}
    while(c>='0'&&c<='9') {x=(x<<1)+(x<<3)+c-'0';c=getchar();}
    return x*f;
}
ll _gcd(ll m, ll n){return n == 0 ? m : _gcd(n, m%n);}
ll _lcm(ll m, ll n){return m*n / _gcd(m, n);}
ll pows(ll base, ll power,ll mod){ll result=1;while(power>0){if(power&1){result=result*base%mod;}power>>=1;base=(base*base)%mod;}return result;}
ll poww(ll base, ll power){ll result=1;while(power>0){if(power&1){result=result*base;}power>>=1;base=(base*base);}return result;}
void init(){ios::sync_with_stdio(0);cin.tie(0),cout.tie(0);}
const int MAX=2e5+10;
int t;
bool cmp(int a,int b){return a>b;}
ll max(ll a,ll b){if(a>b)return a;else return b;}
ll min(ll a,ll b){if(a>b)return b;else return a;}


void solve(){
    int a[MAX];int n;cin>>n;int dp[50005][2][2];
    memset(dp,0,sizeof dp);
    for(int i=1;i<=n;i++){cin>>a[i];}int k;cin>>k;
    for(int i=1;i<=n;i++) a[i]-=k;
    dp[1][0][1]=1;
    if(n==1) {cout<<1<=0){
            dp[i][1][1]=dp[i-1][0][1]+1;
            if(a[i-2]+a[i-1]+a[i]>=0){
                dp[i][1][1]=max(dp[i][1][1],dp[i-1][1][1]+1);
            }
        }
    }
    int ans=0;
    for(int i=1;i<=n;i++){
        for(int j=0;j<2;j++){
            for(int z=0;z<2;z++){
                ans=max(ans,dp[i][j][z]);
            }
        }
    }
    cout<>t;
    while(t--) solve();
//    solve();
    return 0;
}

E. Lexicographically Small Enough 题目：

思路分析：

给出二个字符串 s1 s2 我们可以交换相邻二个字符要让s1调整到字典序严格小于s2的最少操作数

首先用最少操作次数使得s[i]

代码实现：

#include 
using namespace std;
typedef long long ll;
typedef unsigned long long  ull;
inline int read() {
    int x=0,f=1; char c=getchar();
    while(c<'0'||c>'9') {if(c=='-') f=-1;c=getchar();}
    while(c>='0'&&c<='9') {x=(x<<1)+(x<<3)+c-'0';c=getchar();}
    return x*f;
}
ll _gcd(ll m, ll n){return n == 0 ? m : _gcd(n, m%n);}
ll _lcm(ll m, ll n){return m*n / _gcd(m, n);}
ll pows(ll base, ll power,ll mod){ll result=1;while(power>0){if(power&1){result=result*base%mod;}power>>=1;base=(base*base)%mod;}return result;}
ll poww(ll base, ll power){ll result=1;while(power>0){if(power&1){result=result*base;}power>>=1;base=(base*base);}return result;}
void init(){ios::sync_with_stdio(0);cin.tie(0),cout.tie(0);}
const int MAX=2e5+10;
int t;
bool cmp(int a,int b){return a>b;}
ll max(ll a,ll b){if(a>b)return a;else return b;}
ll min(ll a,ll b){if(a>b)return b;else return a;}

int n,sum[MAX];
queuequ[30];
string s1,s2;
void add(int x,int d){for(int i=x;i<=n;i+=(i&(-i))){sum[i]+=d;}}
int get_sum(int x){
    int ans=0;
    for(int i=x;i;i-=(i&(-i))){ans+=sum[i];}
    return ans;
}
void updata(ll &x,ll y){
    if(x==-1) x=y;
    else x=min(x,y);
}
void solve(){
    cin>>n;cin>>s1;cin>>s2;s1=" "+s1;s2=" "+s2;
    for(int i=0;i<=25;i++) {while(!qu[i].empty()) qu[i].pop();}
    for(int i=1;i<=n;i++){qu[s1[i]-'a'].push(i);sum[i]=0;}
    for(int i=1;i<=n;i++){add(i,1);}
    ll ans=-1;ll ans2=0;
    for(int i=1;i<=n;i++){
        int num=n+1;int mx=s2[i]-'a';
        for(int j=0;j>t;
    while(t--) solve();
//    solve();
    return 0;
}

Good Bye 2021: 2022 is NEAR

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