CF1328

CF1328A

如果 \(a\) 是 \(b\) 的倍数 输出0

否则二分答案

点击查看代码

#include<bits/stdc++.h>
using namespace std;
#define inl inline
#define ll long long 
#define endl '\n'
#define int ll
const int N=3e6+5;
const int M=1e5+5;
const int inf=0x3f3f3f3f3f3f3f3f;
const int base=131;
const int mod=1e9+7;
inl 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^48);c=getchar();}
    return x*f;
}
inl void write(int x){
    if(x<0)putchar('-'),x=-x;
    if(x>9)write(x/10);
    putchar(x%10+'0');
}
inl void writei(int x){write(x);putchar(' ');}
inl void writel(int x){write(x);putchar(endl);}
int t,n,m,a,b;
signed main(){
    t=read();
    while(t--){
        a=read();b=read();
        if(!(a%b)){writel(0);continue;}
        int l=0,r=1e9,ans;
        while(l<=r){
            int mid=l+r>>1;
            if(mid*b>a)ans=mid*b-a,r=mid-1;
            else l=mid+1;
        }
        writel(ans);
    }
    return 0;
}

CF1328B

考虑两个 \(b\) 位置

bb

bab bba

baab baba bbaa

发现数量为1+2+...+x

可以二分第一个b位置 第二个就是剩下的数量

点击查看代码

#include<bits/stdc++.h>
using namespace std;
#define inl inline
#define ll long long 
#define endl '\n'
#define int ll
const int N=3e6+5;
const int M=1e5+5;
const int inf=0x3f3f3f3f3f3f3f3f;
const int base=131;
const int mod=1e9+7;
inl 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^48);c=getchar();}
    return x*f;
}
inl void write(int x){
    if(x<0)putchar('-'),x=-x;
    if(x>9)write(x/10);
    putchar(x%10+'0');
}
inl void writei(int x){write(x);putchar(' ');}
inl void writel(int x){write(x);putchar(endl);}
int t,n,k,a,b;
signed main(){
    t=read();
    while(t--){
        n=read();k=read();
        int l=0,r=n-1,ans;
        while(l<=r){
            int mid=l+r>>1;
            if(mid*(mid+1)/2<k)ans=mid,l=mid+1;
            else r=mid-1;
        }
        int lp=n-ans-1,rp=n-(k-ans*(ans+1)/2)+1;
        for(int i=1;i<=n;i++)putchar(i==lp||i==rp?'b':'a');putchar(endl);
    }
    return 0;
}

CF1328C

贪心。

前面尽量均摊 出现1之后就尽量让a为0 这样最优

点击查看代码

#include<bits/stdc++.h>
using namespace std;
#define inl inline
#define ll long long 
#define endl '\n'
#define int ll
const int N=3e6+5;
const int M=1e5+5;
const int inf=0x3f3f3f3f3f3f3f3f;
const int base=131;
const int mod=1e9+7;
inl 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^48);c=getchar();}
    return x*f;
}
inl void write(int x){
    if(x<0)putchar('-'),x=-x;
    if(x>9)write(x/10);
    putchar(x%10+'0');
}
inl void writei(int x){write(x);putchar(' ');}
inl void writel(int x){write(x);putchar(endl);}
int t,n,k;
string a,b,c;
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0),cout.tie(0);
    cin>>t;
    while(t--){
        cin>>n>>c;
        a.clear(),b.clear();
        int flag=1;
        for(auto i:c){
            if(flag){
                if(i=='2')a+='1',b+='1';
                if(i=='0')a+='0',b+='0';
                if(i=='1')a+='1',b+='0',flag=0;
            }else{
                if(i=='2')a+='0',b+='2';
                if(i=='0')a+='0',b+='0';
                if(i=='1')a+='0',b+='1';
            }
        }
        cout<<a<<endl<<b<<endl;
    }
    return 0;
}

CF1328D

贪心。

都一样全填1.

n偶数121212...

n奇数找是否有相邻的相同 这样让相邻11 其余1212

否则123123...

点击查看代码

#include<bits/stdc++.h>
using namespace std;
#define inl inline
#define ll long long 
#define endl '\n'
#define int ll
const int N=3e6+5;
const int M=1e5+5;
const int inf=0x3f3f3f3f3f3f3f3f;
const int base=131;
const int mod=1e9+7;
inl 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^48);c=getchar();}
    return x*f;
}
inl void write(int x){
    if(x<0)putchar('-'),x=-x;
    if(x>9)write(x/10);
    putchar(x%10+'0');
}
inl void writei(int x){write(x);putchar(' ');}
inl void writel(int x){write(x);putchar(endl);}
int t,n,a[N],b[N];
signed main(){
    t=read();
    while(t--){
        n=read();
        for(int i=1;i<=n;i++)a[i]=b[i]=read();
        sort(b+1,b+n+1);
        int m=unique(b+1,b+n+1)-b-1;
        if(m==1){
            writel(1);
            for(int i=1;i<=n;i++)writei(1);putchar(endl);
            continue;
        }
        if(!(n&1)){
            writel(2);
            for(int i=1;i<=n;i++)writei(i&1?1:2);putchar(endl);
            continue;
        }
        for(int i=1;i<=n;i++)a[i]=lower_bound(b+1,b+m+1,a[i])-b;
        int flag=a[1]==a[n],pos=flag?1:0;
        for(int i=2;i<=n;i++){
            if(flag)break;
            flag|=a[i]==a[i-1];
            pos=flag?i:0;
        }
        if(flag){
            writel(2);
            for(int cnt=1,i=pos;cnt<=n;cnt++,i++){
                if(i>n)i-=n;
                a[i]=cnt&1?1:2;
            }
            for(int i=1;i<=n;i++)
                writei(a[i]);putchar(endl);
        }else{
            writel(3);
            for(int i=1;i<=n-(n%3);i++)
                writei(((i-1)%3)+1);
            if(n%3==1)writei(2);
            if(n%3==2)writei(1),writei(2);
            putchar(endl);
        }
    }
    return 0;
}

CF1328E

发现只要保证这些点父亲在一条链上即可

暴力lca判断即可

点击查看代码

#include<bits/stdc++.h>
using namespace std;
#define inl inline
#define ll long long 
#define endl '\n'
#define int ll
const int N=3e6+5;
const int M=1e5+5;
const int inf=0x3f3f3f3f3f3f3f3f;
const int base=131;
const int mod=1e9+7;
inl 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^48);c=getchar();}
    return x*f;
}
inl void write(int x){
    if(x<0)putchar('-'),x=-x;
    if(x>9)write(x/10);
    putchar(x%10+'0');
}
inl void writei(int x){write(x);putchar(' ');}
inl void writel(int x){write(x);putchar(endl);}
int n,m,k,head[N],to[N],nxt[N],cnt;
inl void add(int u,int v){
    nxt[++cnt]=head[u];
    to[cnt]=v;
    head[u]=cnt;
}
int dep[N],f[N][22];
vector<int>v;
inl void dfs(int x,int fa){
    f[x][0]=fa;dep[x]=dep[fa]+1;
    for(int i=1;i<=20;i++)f[x][i]=f[f[x][i-1]][i-1];
    for(int i=head[x];i;i=nxt[i]){
        int y=to[i];
        if(y==fa)continue;
        dfs(y,x);
    }
}
inl bool find(int x,int y){
    for(int i=20;~i;i--){
        if(dep[f[x][i]]<dep[y])continue;
        x=f[x][i];
    }
    return x==y;
}
inl bool cmp(int a,int b){return dep[a]<dep[b];}
signed main(){
    n=read();m=read();
    for(int i=1;i<=n-1;i++){
        int u=read(),v=read();
        add(u,v);add(v,u);
    }dfs(1,0);
    while(m--){
        k=read();int ans=1;v.clear();
        for(int i=1;i<=k;i++)v.push_back(f[read()][0]);
        sort(v.begin(),v.end(),cmp);
        for(int i=1;i<k;i++)ans&=find(v[i],v[i-1]);
        puts(ans?"YES":"NO");
    }
    return 0;
}

CF1328F

每次一定让其全变成原序列中某个数最优

排序后维护前后缀和 那么枚举每个位置 可以 \(O(1)\) 求出答案

点击查看代码

#include<bits/stdc++.h>
using namespace std;
#define inl inline
#define ll long long 
#define endl '\n'
#define int ll
const int N=4e5+5;
const int M=1e5+5;
const int inf=0x3f3f3f3f3f3f3f3f;
const int base=131;
const int mod=1e9+7;
inl 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^48);c=getchar();}
    return x*f;
}
inl void write(int x){
    if(x<0)putchar('-'),x=-x;
    if(x>9)write(x/10);
    putchar(x%10+'0');
}
inl void writei(int x){write(x);putchar(' ');}
inl void writel(int x){write(x);putchar(endl);}
int n,k,a[N],pre[N],suf[N],ans=inf,res;
signed main(){
    n=read();k=read();
    for(int i=1;i<=n;i++)a[i]=read();
    sort(a+1,a+n+1);
    int cnt=1;
    for(int i=2;i<=n;i++){
        if(a[i]^a[i-1])res=max(res,cnt),cnt=1;
        else cnt++; 
    }res=max(res,cnt);
    if(res>=k){puts("0");return 0;}
    for(int i=1;i<=n;i++)pre[i]=pre[i-1]+a[i];
    for(int i=n;i;i--)suf[i]=suf[i+1]+a[i];
    for(int i=1;i<=n;i++){
        if(i>=k)ans=min(ans,i*a[i]-pre[i]-(i-k));
        else ans=min(ans,(2*i-n-1)*a[i]-pre[i]+suf[i]-(n-k));
        if(n-i+1>=k)ans=min(ans,suf[i]-(n-i+1)*a[i]-(n-i+1-k));
        else ans=min(ans,(2*i-n-1)*a[i]-pre[i]+suf[i]-(n-k));
    }
    writel(ans);
    return 0;
}